Second in-class problem from 2012-07-05

1.92M NF₃(g) + 1.63M O₂(g) → NOF₃(g) K_eq=1.74x10⁹, find all [ ]_eq

	2 NF3(g) +	O2(g) ⇄	2 NOF3(g)
[ ]initial	1.92 M	1.63 M	0 M
Δ [ ]	- 2x M	- x M	+ 2x M
[ ]equilibrium	(1.92 – 2x) M	(1.63 – 2x) M	2x M

Plugging in to the equilibrium constant expression, this would get ugly pretty quickly. Let's try to simplify it with an approximation. Since K_eq is quite product-favored, we can probably assume that the limiting reagent is essentially used up to form (almost) the theoretical yield of product. From the balanced chemical equation, we need 2 NF₃(g) for each O₂(g), but we didn't start with twice as much NF₃(g), so that must be the limiting reactant. For every mole of NF₃(g) that reacts, 1 mol of NOF₃(g) is formed, so the equilibrium concentration of NOF₃(g) will be (just a tiny bit less than) 1.92M. To make (just a tiny bit less than) 1.92M NOF₃(g), we need half that amount of O₂(g), so to reach equilibrium we will consume (very close to) 0.96M O₂(g), leaving (1.63M – 0.96M = 0.67M) of O₂(g) at equilibrium.

{NOTE: If you prefer, you could explicitly calculate this out using our standard 4-step process for limiting reactants/theoretical yields by assuming a volume. I'd assume 1L to make the calculations easier.}

We can now start with a fresh table to organize our information...

	2 NF3(g) +	O2(g) ⇄	2 NOF3(g)
[ ]equilibrium	“a tiny bit”	0.67 M	1.92 M

Now we can plug into the equilibrium constant expression:

Solving, “a tiny bit” = 5.62x10^-5M. Make sure you always CHECK YOUR ASSUMPTIONS! In this case, yes, 5.62x10^-5 is indeed very small compared to 1.92, so the assumption is valid. One of the nice things about equilibrium problems is that you can always plug your results back in to the equilibrium constant expression and see if you get an answer that's close to what's given in the problem. In this case...

(1.92)² / {5.62x10^-5)²(0.67) = 1.742x10⁹

So...

[NF₃]_eq = 5.62x10^-5 M

[O₂]_eq = 0.67M

[NOF₃]_eq = 1.92M