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January 26, 2015

January 26, 2015

Posted by **Kat** on Wednesday, May 25, 2011 at 7:51am.

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**, Wednesday, May 25, 2011 at 12:31pmKat--I 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......2.235-2x

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 2.235-x term that it isn't noticeable. SO, here is the only way I see to do it. Look at k. With a value of 1.2E-10 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).

1.2E-10 = (O)^2/(O2)

1.2E-10 = (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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