# How do you determine the equilibrium constant K_c for the reaction X + 2Y rightleftharpoons Z? The equilibrium concentrations are "0.216 M" for Z, "0.06 M" for X, and "0.12 M" for Y.

Aug 24, 2017

${K}_{c} = 250$

#### Explanation:

The equilibrium constant is simply a ratio between two multiplications.

In the numerator, you have the multiplication of the equilibrium product concentrations raised to the power of the products' respective stoichiometric coefficients.

In the denominator, you have the multiplication of the equilibrium reactant concentrations raised to the power of the reactants' respective stoichiometric coefficients.

In your case, the equilibrium reaction looks like this

$\text{X" + 2"Y" rightleftharpoons "Z}$

which means that you have

• $\text{Products:}$

$\text{Z " -> " no coefficient = coefficient of 1}$

• $\text{Reactants:}$

$\text{X " -> " no coefficient = coefficient of 1}$

$\text{Y " -> " coefficient of 2}$

By definition, the equilibrium constant for this reaction will be

${K}_{c} = \left({\left[\text{Z"]^1)/(["X"]^1 * ["Y}\right]}^{2}\right)$

which is equivalent to

${K}_{c} = \left({\left[\text{Z"])/(["X"] * ["Y}\right]}^{2}\right)$

Plug in the values you have for the equilibrium concentrations of the three chemical species to find the value of the equilibrium constant--I'll leave the answer without added units

${K}_{c} = \frac{0.216}{0.06 \cdot {0.12}^{2}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{250}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the equilibrium concentration of $\text{X}$.