Chemistry
posted by David on .
At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.100 M and [NO] = 0.500 M.
N2(g) + O2(g) <> 2NO(g)
If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is reestablished?
Kc = 0.800 M/[0.100M][0.100M] = 80 M?

the extra NO being added threw me off..
or is it supposed to be like Kc = 0.800M^2/[0.100M][0.100M] = 64 M? 
No and no.
First, since the values quoted are equilibrium values, we need to calculate Kc (by the way, 80 is not an equilibrium value).
............N2 + O2 ==> 2NO
equil.......0.1..0.1.....0.5
Kc = (NO)^2/(N2)(O2)
Kc = (0.5)^2/(0.1)(0.1)
Kc = 25 is what I have.
Then we set up another ICE chart.
............N2 + O2 ==> 2NO
initial.....0.1...0.1....0.8
change.....+x.....+x.....2x
equil......0.1+x..0.1+x...0.8x
Kc = 25 = (NO)^2/(N2)(O2)
25 = ((0.82x)^2/(0.1+x)(0.1x)
Solve for x.
By the way, you need NOT go through a quadratic if you notice that you can take the square root of both sides. 
why does the equil end up being 0.8x instead of 0.8  2x? and i solved for x but it gave me 2 different answers did I do something wrong?

First, it wasn't 2x because I made a typo.It's obvious from the initial value and the change value that it SHOULD be 0.8002x and, in fact, I wrote (0.8002x)^2 when I substituted into the K expression.
You ALWAYS get two answers when you solve a quadratic. Most of the time it is obvious which is the right one and which is the wrong one. For example, I would expect, although I didn't go through the math, if you take the two x values you have, multiply by 2 and subtract from 0.800 that one will be a positive number and one a negative number. The one giving a negative number when subtracted from 0.800 will be the one to throw away since you can't have negative concns. 
Cngkreng jkt .

I don't know how to do the last algebra part that DrBob22 did not explain.
I have not taken algebra since 2 years ago it would have been helpful to see...
I haven't found out the math part anywhere! 
How do we construct an ICE chart for the concns of the dissociated ions to form a precipitate so we can calculate the final concns of ions in solution?