I need help on finding the initial equilibrium concentrations

Question

A+2B=3c

the reaction mixture is samples prior to the equilibrium being reached. It is determined that the reaction must proceed to the left for equilibrium to be reached. When the reaction mixer is sampled at equilibrium the concentrations are A= 0.267mol L-1 B= 0.336mol L-1 C= 0.173mol L-1

what were the concentrations of A,B and C when the reaction mixture was sampled prior to the equilibrium?

Answer 1

Here is how I might approach such a problem. Use an ICE table (ICE box) to set up the variables.

A + 2B < === > 3C

?.......?................>.......Initial

-x......-2x............+3x.....Change

0.267...0.336.....0.173...Equilibrium

Thus, the ? for the initial concentration can be stated as

[A] = 0.267 + x

[B] = 0.336 + 2x

[C] = 0.173 - 3x

At equilibrium:

Keq = [C]³/[A][B]² = (0.173)³/(0.267)(0.336)² = 0.1718

0.1718 = (0.173 - 3x)³/(0.267 + x)(0.336 + 2x)²

Solve for x and then to find initial concentrations, add x to 0.267 and to 0.336 to get A and B respectively, and subtract x from 0.173 to obtain C.

The math is something I'm not comfortable performing, so I've left that to you.