chemistry
posted by Brunette .
The equilibrium constant Kc for the reaction
C <> D + E
is 7.90 * 10^5. The initial composition of the reaction mixture is [C]=[D]=[E]=1.10*10^3. What is the equilibrium concentrations of C, D, and E?
C=?
D=?
E=?
chemistry  DrBob222, Monday, November 5, 2012 at 12:21am
K = 7.9E5 = (D)(E)/(C)
Qc = (0.0011)(0.0011)/(0.0011) = 0.0011
Qc > Kc; therefore, products are too large and reactants too small. The reaction most go to the left to reach equilibrium.
..........C ==> D ...+... E
I......0.0011..0.0011..0.0011
C.........x.....x.......x
E...0.0011+x..0.0011x.0.0011x
Substitute the equilibrium line from the ICE chart into Kc expression and solve for x, then 0.0011+x and 0.0011x.
I got this when I used the quadratic formula(the answer to the quad form was 1.54*10^3):
D: 0.0011 1.5*10^3= 0.00044
E: 0.0011 1.5*10^3= 0.00044
C: 0.0011 + 1.54*10^3 = 0.00264
I got the wrong answers. It is saying I solved the equation correctly, but I need to use the other root of the equation in the ICE chart. What do they mean by that?

That's exactly right. Before I posted the answer for you last night I worked the problem out completely. You KNOW 1.5E3 can't be right. Why? Because 1.5E3 is more than you started with (1.10E3) so you use the other root of the quadratic equation. (So I used the other root and it worked okI obtained a reasonable answer). (You know when you solve the quadratic of
[b +/ sqrt(b^24ac)/2a].
So use the other root of the quadratic and things will work out ok. I think the other root gives x = about 0.00072 (I don't have notesthat's the number I remember so confirm that). 
Oh! I see what you're saying! See, that's what I thought it meant, but I just wanted to make sure. I have a tendancy to second guess myself when it comes to chemistry.

Don't we all.