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2NF3(g) yields N2(g) + 3F2(g)

When 2.06 mol of NF3 is placed in a 2.00L container and allowed to come to equilibrium at 800K, the mixture is found to contain 0.0227 mol of N2. What is the value of Kp at this temp?

1.91 X 10-3
1.73 X 10-6 (think this is it)
4.43 X 10-7
1.83 X 10-3

  • Chemistry - ,

    I didn't get that. Post your work and I'll look for the error.

  • Chemistry - ,

    I think I tried to do the ICE and got...

    2.06/2.00=1.03 M
    0.0227/2.00=0.0114 M

    but got lost after that and ended up with 1.73 X 10-6...I am really struggling on these problems...

  • Chemistry - ,

    I think you started out wrong. The problem gives as Kp and you are calculating, by mols/L, the concn. You put pressures in Kp and concns in Kc.

    ........2NF3 ==> N2+ 3F2
    C.,.....-2x.... .x....3x

    The problem tells you that x(N2) is 0.0227 mols. That makes F2 = 3*0.0227 and mols NF3 = 2.06- (2*0.0227)

    Now convert 0.0227 to pressure using PV = nRT and do the same to find pressure of F2 and NF3. The substitute into Kp expression and solve for Kp.
    My answer was 0.00190

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