# What is the equilibrium constant of PCl_5 (g) -> PCl_3(g) + Cl_2(g)?

##### 1 Answer
Dec 14, 2016

Here's what I got.

#### Explanation:

Because you're dealing with a reaction that features gases, you can actually write two expressions for the equilibrium constant.

One expression will feature molar concentrations and it will give you the value of ${K}_{c}$, while the other will feature partial pressures and give you the value of ${K}_{p}$.

So, your equilibrium reaction looks like this

${\text{PCl"_ (5(g)) rightleftharpoons "PCl"_ (3(g)) + "Cl}}_{2 \left(g\right)}$

Now, the expression for ${K}_{c}$ is given by the ratio that exists between the equilibrium concentrations of the products raised to the power of their respective stoichiometric coefficients and the equilibrium concentrations of the reactants raised to the power of their respective stoichiometric coefficients.

In this case, you have

K_c = (["PCl"_3] * ["Cl"_2])/(["PCl"_5]

The expression for ${K}_{p}$ is given by the ratio that exists between the equilibrium partial pressures of the products raised to the power of their respective stoichiometric coefficients and the equilibrium partial pressures of the reactants raised to the power of their respective stoichiometric coefficients.

In this case, you have

${K}_{p} = \left(\left({\text{PCl"_3) * ("Cl"_ 2))/(("PCl}}_{5}\right)\right)$