Question

Haber process at 127C:

1N2 + 3H2 -----> 2 NH3
[NH3] @ equilibrium= 0.031M
[N2] @ initial =0.85 M
[H2] @ initial =0.0031 M
Calc Keq for the reaction. This is how I did it:
1N2 + 3H2 -----> 2 NH3
i:0.85 0.0031 0
c: -X -3X +2X
e: 0.85-X 0.0031-3X 0.031=2X
X=0.0155
Keq= ([0.031]^2/([0.8345]*[-0.0434]^3)

I am not sure why I am getting a negative value for the eq. [H2]????
What am I doing wrong? Thank you.
Holly

Answer 1

You aren't doing anything wrong. You could have the initial and equilibrium amounts for H2 and NH3 reversed. Or the problem could have a misprint in it. Your work looks good to me and I might add that it is being observant to know that something must be wrong with the answer because of the negative sign.