Solve Equilibrium for Chemical Reaction: 2A <-> B + C

[Hollysmoke]

Homework Statement

The following reaction initially contains [A]o = 0.2150 M and o = [C]o = 0.4150 M. If the equilibrium constant for the reaction is Keq = 2.03, what are the concentrations of reactants and products when the reaction has achieved equilibrium? What is the value of deltaG for the reaction at 25oC?

2 A <-> B + C

Homework Equations

I am stuck as on how to proceed with the math. I'm not sure if I am doing it correctly or not. Could someone please take a look?

The Attempt at a Solution

Initial: 2A B C
0.2150M 0.4150M 0.4150M

Change: +x -x -x
Since Q>Kc, as Qc = [C]/[A]^2

Equil. 0.2150+x 0.4150-x 0.4150-x

I'm stuck on the whole rearranging and solving for x bit. I know I'm supposed to use quadratic formula.

 

[chemisttree]

Substitute the values you know for the equilibrium concentration of A, B and C (A=0.2150+x, B=0.4150-x, C=0.4150-x) into the expression for equilibrium and solve it.

Keq=2.03=[(0.4150-x)^2/(0.2150+x)^2] and expand it out, multiply both sides by (A+x)^2, do the algebra until the expression for Keq is in the form:

0 = ax^2 + bx + c (note that here the terms a, b and c are not the same as A, B and C given in the problem)

Hint: The first step should look like this:

2.03*(0.2150+x)^2 = (0.4150-x)^2, after multiplying both sides by (0.2150+x)^2