Question

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 = 7.9E-5 = (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..0.0011-x.0.0011-x

Substitute the equilibrium line from the ICE chart into Kc expression and solve for x, then 0.0011+x and 0.0011-x.


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?

Answer 1

That's exactly right. Before I posted the answer for you last night I worked the problem out completely. You KNOW 1.5E-3 can't be right. Why? Because 1.5E-3 is more than you started with (1.10E-3) so you use the other root of the quadratic equation. (So I used the other root and it worked ok--I obtained a reasonable answer). (You know when you solve the quadratic of
[-b +/- sqrt(b^2-4ac)/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 notes--that's the number I remember so confirm that).

Answer 2

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.

Answer 3

Don't we all.