# Question d78b6

Jul 19, 2017

["N"_ 2"O"_ 4] = "0.003 mol L"^(-1)

#### Explanation:

You know that a certain temperature, you have

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

with an equilibrium constant of

${K}_{e q} = 1.15$

Now, by definition, the equilibrium constant for this reaction takes this form--don't forget that the expression of the equilibrium constant takes into account the stoichiometric coefficients of the chemical species involved in the reaction!

${K}_{e q} = \left({\left[{\text{N"_2"O"_4])/(["NO}}_{2}\right]}^{\textcolor{red}{2}}\right)$

This means that you have

${\left[{\text{N"_ 2"O"_ 4] = K_(eq) * ["NO}}_{2}\right]}^{\textcolor{red}{2}}$

Plug in your values to find--since you didn't get units for ${K}_{e q}$, I won't use units for the concentration of nitrogen dioxide, but I will add units for the concentration of dinitrogen tetroxide!

["N"_ 2"O"_ 4] = 1.15 * (0.05)^color(red)(2) = color(darkgreen)(ul(color(black)("0.003 mol L"^(-1))))#

The answer is rounded to one significant figure, the number of sig figs you have for the equilibrium concentration of nitrogen dioxide.