Asked by Holly
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
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
Answers
Answered by
DrBob222
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.
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