Asked by Robby

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

Answers

Answered by DrBob222
I didn't get that. Post your work and I'll look for the error.
Answered by Robby
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...
Answered by DrBob222
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
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions