Chemistry
posted by Kat .
The reaction O2 ⇄ 2 O has a K=1.2x1010. If a 1.7 L container holds 3.8 moles
of O atoms (and nothing else), what will the concentration of O be when
equilibrium is reached?
So far, I have 1.2x10^10 = 2.2^2 / 0
but I don't know how to do the rest. When it says "when equilibrium is reached" does it mean that the left side and the right side has to match? Help!!

Chemistry 
DrBob222
KatI have struggled with this problem. No, equilibrium doesn't mean the left side and the right side are equal. It means that the net amount of O and O2 are not changing with time. Set up would go like this.
...........O2 > 2O
initial....0......3.8/1.7 = 2.235M
change....+x......2x
equil.....+x......2.2352x
Then K = (O)^2/(O2) and substitute the equil values from the ICE chart above. BUT this equation has no solution; either that or I didn't find it. In some instances the equation can be reversed (for which k is k' = 1/k), especially with small numbers and I did find a solution for that reverse equation but when I plug that value into the expression for k I don't get k which means the solution is not correct. The problem with the quadratic is that the small value of k leads to such a small change in the 2.235x term that it isn't noticeable. SO, here is the only way I see to do it. Look at k. With a value of 1.2E10 it means that equilibrium lies far to the left; i.e., as O2 and not O. Therefore, with such a small number, we simply say that the equilibrium value of O2 will be 2.235 (essentially ALL of the O atoms combine to form O2). Then we plug that into the K expression and solve for (O).
1.2E10 = (O)^2/(O2)
1.2E10 = (x^2)/2.235
and solve for x and you arrive at a very small number for x which you expect. Check my thinking. If you find this is not the way to do it OR that the quadratic equations do in fact have a reasonable solution please post that to the site. Bob Pursley may look at this and offer his assessment, too.
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