9.2g of dinitrogen tetroxide is taken in a closed 1L vessel and heated till the following equilibrium is reached. At equilibrium, 50% of the "N"_2"O"_4(g) is dissociated. What is the equilibrium constant (in mol/liter) ?

${\text{N"_2"O"_4(g) rightleftharpoons 2"NO}}_{2} \left(g\right)$

Nov 8, 2016

${K}_{c} = {\text{0.20 mol L}}^{- 1}$

Explanation:

The equilibrium reaction given to you looks like this

${\text{N"_ 2"O"_ (4(g)) rightleftharpoons color(blue)(2)"NO}}_{2 \left(g\right)}$

Your first goal here is to figure out the initial concentration of the dinitrogen tetroxide. You know that because the vessel has a volume of $\text{1.0 L}$, you can practically treat number of moles and concentration interchangeably.

Use the molar mass of dinitrogen tetroxide to find the number of moles present in the sample

9.2 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"O"_4)/(92.011color(red)(cancel(color(black)("g")))) ~~ "0.100 moles N"_2"O"_4

This means that the initial concentration of dinitrogen tetroxide is equal to ${\text{0.100 mol L}}^{- 1}$.

Now, the problem tells you that at equilibrium, 50% of the added dinitrogen tetroxide is dissociated. This basically means that half of the number of molecules of dinitrogen tetroxide will dissociate to form nitrogen dioxide.

So, if half of the initial concentration gets converted to nitrogen dioxide, you can say that the equilibrium concentration of dinitrogen tetroxide will be

["N"_ 2"O"_ 4]_ "eq" = 1/2 * "0.100 mol L"^(-1) = "0.0500 mol L"^(-1)

Notice that for every molecule of dinitrogen tetroxide that dissociates, $\textcolor{b l u e}{2}$ molecules of nitrogen dioxide are produced.

This means that the equilibrium concentration of nitrogen dioxide will be twice the concentration of dinitrogen tetroxide that dissociated.

["NO"_ 2]_ "eq" = color(blue)(2) * "0.0500 mol L"^(-1) = "0.100 mol L"^(-1)

By definition, the equilibrium constant for this reaction will be

K_c = (["NO"_ 2]_ "eq"^color(blue)(2))/(["N"_ 2"O"_ 4]_ "eq")

Plug in your values to find

K_c = ("0.100 mol L"^(-1))^color(blue)(2)/("0.0500 mol L"^(-1)) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.20 mol L"^(-1))color(white)(a/a)|)))

The answer is rounded to two sig figs.