# What is the equilibrium constant for the reaction #2"CH"_ 2"Cl" _ (2(g)) rightleftharpoons "CH"_ (4(g)) + "CCl"_ (4(g))# ?

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Given that, at equilibrium, the vessel contains:

#["CH"_2"Cl"_2] = "0.021 M"#

#["CH"_4] = "0.2 M"#

#["CCl"_4] = "0.2 M"#

Given that, at equilibrium, the vessel contains:

##### 1 Answer

#### Explanation:

The equilibrium reaction given to you looks like this

#2"CH"_ 2"Cl" _ (2(g)) rightleftharpoons "CH"_ (4(g)) + "CCl"_ (4(g))#

Now, the equilibrium constant takes into account the *equilibrium concentrations* of the products and of the reactants **and** the stoichiometric coefficients that correspond to them.

More specifically, the equilibrium constant is calculated by **dividing** the product of the equilibrium concentrations of the **products** raised to the power of their respective stoichiometric coefficients by the product of the equilibrium concentrations of the **reactants** raised to the power of their respective stoichiometric coefficients.

In this case, you have

#"CH"_4 -> "no coefficient = coefficient of 1"# #"CCl"_4 -> "no coefficient = coefficient of 1"# #"CH"_2"Cl"_2 -> "coefficient of 2"#

The equilibrium constant will take the form

#K_c = (["CH"_4]^1 * ["CCl"_4]^1)/(["CH"_2"Cl"_2]^2)#

which is equivalent to

#K_c= (["CH"_4] * ["CCl"_4])/(["CH"_2"Cl"_2]^2)#

To find the actual value of the equilibrium constant at the given temperature, plug in the values you have for the equilibrium concentrations of the three chemical species

#K_c = (0.2 color(red)(cancel(color(black)("M"))) * 0.2 color(red)(cancel(color(black)("M"))))/(0.021^2 color(red)(cancel(color(black)("M"^2)))) = 90#

The answer is rounded to one **significant figure**.