Consider the following reaction. A(aq) <---> 2B(aq) Kc = 5.17 x 10^-6

Question

I have been trying to figure out how to do this problem. I am not that good at ICE tables. Consider the following reaction. A(aq) <--> 2B(aq) Kc = 5.17 x 10^-6 at 500K. If a 5.10 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

In A(aq) <==> 2B(aq), you can see that for every 1 unit of A that disappears, 2 units of B appear. This will be reflected in the ICE table, as follows:

A(aq) <=====> 2B(aq)

5.10..................0.........Initial

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

5.10-x.............2x........Equilibrium

Kc = [B]²/[A]

5.17x10^-6 = (2x)²/5.10-x

5.17x10-6 = 4x2/5.10-x

2.637x10-5 - 5.17x10^-6x = 4x²

4x² + 5.17x10-⁶x - 2.62x10^-5 = 0

Solving for x using the quadratic we obtain

x = 0.00255 M

At equilibrium [B] = 2 x 0.00255 = 0.0051 M