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

Question

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

Answer 1

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

Answer 2

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...

Answer 3

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
I......2.06mols...0....0
C.,.....-2x.... .x....3x
E.....2.06-x......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