Equilibrium Problem Emergency?

Question

This equilibrium problem has been killing me for quite a while. And I only have about an hour or two to figure it out. Any help appreciated greatly! Thanks


A + B <-> 2C

The above reaction has an equilibrium constant Kc = 115 at a certain temperature. In a particular experiment, 3.191 mol A, 3.191 mol B, and 3.191 mol C were added to a 1.500 L flask. Calculate the equilibrium concentrations of all species.

Answer 1

The equilibrium eqtn is
4[Cc]^2 / [Cb][Ca] = 115
If you stuck 3.19 in this equation, you would get a constant of about 4, so it looks like C is being made at the expense of A and B. Suppose, at equilibrium, we have lost x moles of A and B.
Then, 4 [3.19+2x]^2 / (3.19-x)^2 = 115.

Just solve, make sure you select the right root.

Answer 2

Initial concentrations:

[A](0) = 2.127 M
[B](0) = 2.127 M
[C](0) = 2.127 M

At equilibrium, 115 = [C]^2 / [A][B]

Since K(c) > 115, there is much more of "C" than A or B at this time.

So, 115 = (2.127 + 2x) / (2.127 - x)(2.127 - x)

x = 2.371 M or 1.900 M
x = 1.900 M since (2.127 - x) > 0

[C] = 5.927 M
[A] = [B] = 0.227 M