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

## Given that, at equilibrium, the vessel contains: ["CH"_2"Cl"_2] = "0.021 M" ["CH"_4] = "0.2 M" ["CCl"_4] = "0.2 M"

Aug 22, 2017

${K}_{c} = 90$

#### Explanation:

The equilibrium reaction given to you looks like this

$2 {\text{CH"_ 2"Cl" _ (2(g)) rightleftharpoons "CH"_ (4(g)) + "CCl}}_{4 \left(g\right)}$

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

• $\text{CH"_4 -> "no coefficient = coefficient of 1}$
• $\text{CCl"_4 -> "no coefficient = coefficient of 1}$
• $\text{CH"_2"Cl"_2 -> "coefficient of 2}$

The equilibrium constant will take the form

${K}_{c} = \left({\left[{\text{CH"_4]^1 * ["CCl"_4]^1)/(["CH"_2"Cl}}_{2}\right]}^{2}\right)$

which is equivalent to

${K}_{c} = \left({\left[{\text{CH"_4] * ["CCl"_4])/(["CH"_2"Cl}}_{2}\right]}^{2}\right)$

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

${K}_{c} = \left(0.2 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{M"))) * 0.2 color(red)(cancel(color(black)("M"))))/(0.021^2 color(red)(cancel(color(black)("M}}^{2}}}}\right) = 90$

The answer is rounded to one significant figure.