Hey guys, I was doing some homework I missed in Chemistry and I need help doing this ...

The section I'm on is called "Reversible Reactions and Equilibrium". If you know how to do this, please help.

There are only two questions but I'm sure I can do the second one by using the first one as an example.



It also says here that:

k_eq = [C]^c X [D]^d/[A]^a X [B]^b

Any help/insight somebody can give is much appreciated.

Ask me this a few months and I would've known the answer. Sorry, I couldn't help you out bud.

Thanks for trying at least.

It took some intense concentration and thinking but I'm pretty sure I got it now.

If you want to check:

The reversible reaction N2(g) + 3H2(g) ? 2NH3(g) produces ammonia, which is a fertilizer. At equilibrium, a 1-L flask contains 0.15 mol H2, 0.25 mol N2, and 0.10 mol NH3. Calculate Keq for the reaction.

keq = [C]^c X [D]^d/[A]^a X [B]^b

In this scenario, [A] is the concentration of N2, [B] is the concentration of H2 and [C] is the concentration of NH3. There is no [D], as no fourth reagent is included. For this equilibrium quotient thing, you always take what is behind the double arrow as the top number, demonstrated earlier in this sentence.
a, b, c stand for the amount of mole there is relative to each other. In this reaction, a is 1, b is 3 and c is 2.

That gives us:

Keq = [NH3]^2 / [N2]^1 * [H2]^3

Now, you have a 1 litre flask. That means that the mol given is instantly also the concentration. Respectively:
0.15 mol/litre H2, 0.25 mol/litre N2, 0.10 mol/litre NH3.

In the formula, this gives us:

Keq = 0.10^2 / (0.25 * 0.15^3)
Keq = 0.01 / (0.25 * 0.003375)
Keq = 0.01 / (0.00084375)
Keq = 11.8518519

..I think, anyway. I might have made a mistake somewhere. This should give you a general idea of how to do this though.
Note that you can never use liquids or solids in this formula, only gases and solutions.

Been SUCH a long time since i've done chemistry and had to remove myself from the class during rate reactions. However looking at your methodology Oranos it looks correct the only thing I would scrutinize would be the written out method:

[b]Keq = 0.10^2 / (0.25 * 0.15^3)
Keq = 0.01 / (0.25 * 0.003375)
Keq = 0.01 / (0.00084375)
Keq = 11.8518519

Otherwise everything looks correct imho from what I remember in chemistry class. Was about to stop and say that the number for nitrogen didn't look right, but it was.