chemistry

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I need help. I have no clue how to find this.
A flask is charged with 1.680 atm of N2O4(g) and 1.220 atm of NO2(g) at 25 degrees celcius. The equilibrium reaction is
N2O4 <-> 2 NO2
After equilibrium is reached, the partial pressure of NO2 is 0.550
(a) What is the equilibrium partial pressure of N2O4?
(b) Calculate the Kp for the reaction

I have tried
1.22^2/ 1.680 = 0.550^2/x
but the Webassign said it was wrong. I have also tried
1.220^2/1.680 = 0.886
0.55^2/x = 0.886
x= 2.9289

  • chemistry -

    Note the correct spelling of Celsius.

    .........N2O4 <-> 2NO2
    I........1.680.....1.220
    C.........-p.......+2p
    E........1.680-p...2p

    The problem tells you that 2p = 0.550. You can calculate p from that.

    That allows you to evaluate 1.680-p

    Then substitute these values into the Kp expression and evaluate Kp.

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