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

##### 1 Answer

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 **partial pressures** and give you the value of

So, your equilibrium reaction looks like this

#"PCl"_ (5(g)) rightleftharpoons "PCl"_ (3(g)) + "Cl"_ (2(g))#

Now, the expression for *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 *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 = (("PCl"_3) * ("Cl"_ 2))/(("PCl"_5))#