2012/07/06

Second in-class problem from 2012-07-05


1.92M NF3(g) + 1.63M O2(g) → NOF3(g) Keq=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 Keq 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


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