NH​3​ is normally encountered as a gas with a pungent odor. It is

Question

formed from hydrogen and nitrogen (equation given below). A
chemist pours a mixture of nitrogen gas and hydrogen gas in a 1.0
litre reaction vessel in 1:3 ratio. The equilibrium constant for the
reaction is 1.0 × 10​^–4, and the equilibrium value of [H​2​(g)] is 0.12 mol/L. Calculate the equilibrium value of [NH​3​].

N​2​(g) + 3 H​2​(g) ⇔ 2 NH​3​(g)

I know this is a late responds, but I would like to hear your explanation to the question above. I have finally gotten the answer to the question above and the correct answer is [NH​3] = 8.314 x 10^-5 mol/L. Thanks for 4 years of help, Dr.bob.

Answer 1

We previously said the answer was 8.3E-4 M for (NH3) given the conditions. What we didn't know was if the equilibrium ratio of N2:H2 was 1:3 if the starting ratio was 1:3. ASSUMING the equilibrium ratio for this reaction is the same as the starting ratio then Kc = 1E-4 = (NH3)^2/(N2)(H2)^3. Plug in 0.12 for (H2) from the problem, plug in 1/3 of that for the N2 (0.12/3 = 0.04), then the ONLY answer you can get for (NH3) is roughly 8.3E-4 M for NH3. Given that the (H2) was fixed @ 0.12M, that no starting material was given for H2 or N2, and no value of (N2) @ equilibrium was given I think we are forced to make the assumption that N2 @ equilibrium is 0.04 and solve that way. What about the assumption? I've never seen that in my career so I looked at the equation A + 2B ==> C and plunged in 0.5M and 1M for A:B. You can go through the math, as I did, and lo and behold, the end value at equilibrium is 1:2 ratio for A:B. Then I changed to 3:6 for different values BUT the ratio still was 1:2 for A:B. The ratio of A:B at equilibrium was lo and behold 1:2. Then I changed to ratio of A:B of 4:6 (2:3 or 1:1.5) and the end value at equilibrium for A:B was NOT 2:3 (1:1.5). It was 1:1.39. I've never seen this argued in any text, internet, discussion, not anywhere. So I'm assuming that this argument holds: i.e., if the ratio of the reactants is the same as the stoichiometry of the reaction THEN the equilibrium ratio is the same ratio for those reactants. That does not hold true if the initial ratio is not the same as the stoichiometric coefficients.
There is another part to this. Since no starting material was given, I asked myself what would that mystery 1:3 mixture have to be to get, at equilibrium, (N2) = 0.04M, (H2) = 0.12M and (NH3) = 8.314E-5 M. That's easier said than done so it makes me think the values of 0.04M and 0.12 were out of the blue and it was easier to say "a mixture in 1:3 ratio......." than to work it out. Anyway you can see what I mean.
..............N2 + 3H2 ==> 2NH3
I..............y........3y............0
C............-x........-3x..........2x
E...........0.04.....0.12.......2x
We know (NH3) = 8.34E-5 M so x must be 4.17E-5
That makes y = (N2) = 0.04 + 0.0000417 = 0.0400417
You can do (H2) initially but you see that makes (N2) = 0.04 @ equilibrium a ridiculous number to start with. Starting molarity for H2 is almost as bad. Again, I think the starting numbers make no sense given the equilibrium values of 0.04 and 0.12. Just my opinion. I hope all of this makes sense.

Answer 2

Thanks again, I understand your explanation.