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

The equilibrium reaction given to you looks like this

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

The equilibrium constant will take the form

which is equivalent to

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

The answer is rounded to one significant figure.

The equilibrium constant (Kc) for the reaction (2\text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g)) is given by:

[K_c = \frac{[\text{CH}_4][\text{CCl}_4]}{[\text{CH}_2\text{Cl}_2]^2}]