# Question 65812

May 20, 2015

So, you're dealing with a generic equilibrium reaction for which you know the initial concentration of the species that take part in said equilibrium.

The first thing you can do is figure out in which direction the equilibrium will shift once the species are added to the vessel. You can do that by calculating the reaction quotient, ${Q}_{c}$, and comparing it with the equilibrium constant, ${K}_{c}$.

SIDE NOTE You're going to have to use the quadratic equation, it will make your life so much easier. You could try using successive approximations, but that is just a painstaking and error-prone process that I suggest you skip it altogether.

Using the initial concentrations of $X$, $Y$, and $X Y$, the reaction quotient wil be

${Q}_{c} = \frac{{\left[X\right]}_{0} \cdot {\left[Y\right]}_{0}}{{\left[X Y\right]}_{0}}$

Q_c = (0.100 * 0.100)/(0.500) = (0.100"^2)/0.500 = 0.0200

Notice that ${Q}_{c} < {K}_{c}$. This tells you that, initially, you have more reactant than products present in the vessel; as a result, the equilibrium will shift to the right, resulting in more products being formed.

This means that you can expect the equilibrium concentrations of the products to be bigger than 0.100 M, and the equilibrium concentration of the reactant to be smaller than 0.500 M.

$\text{ "XY_((aq)) rightleftharpoons " "X_((aq)) " "+ " } {Y}_{\left(a q\right)}$
I...0.500................0.100.................0.100
C....(-x)....................(+x).....................(+x)
E...0.500-x...........0.100+x.............0.100+x

You know that the equyilibrium constant is equal to

${K}_{c} = \frac{\left[X\right] \cdot \left[Y\right]}{\left[X Y\right]} = \frac{\left(0.100 + x\right) \cdot \left(0.100 + x\right)}{0.500 - x}$

K_c = ((0.100 + x)"^2)/(0500-x) = 0.150#

This is equivalent to

${x}^{2} + 0.35 x - 0.065 = 0$

This equation will produce two values for $x$, a positive one and a negative one. Right off the bat, the negative one is eliminated because $x$ represents the concentration of $X$ and $Y$.

As a result, the solution will be

$x = 0.134$.

This means that the equilibrium concentrations for your species will be

$\left[X Y\right] = 0.500 - 0.134 = \textcolor{g r e e n}{\text{0.366 M}}$
$\left[X\right] = 0.100 + 0.134 = \textcolor{g r e e n}{\text{0.234 M}}$
$\left[Y\right\} = 0.100 + 0.134 = \textcolor{g r e e n}{\text{0.234 M}}$