Question

Question #d78b6

Answer 1

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

Explanation:

You know that a certain temperature, you have

`color(red)(2)"NO"_ (2(g)) rightleftharpoons "N"_ 2"O"_ (4(g))`

with an equilibrium constant of

`K_(eq) = 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_(eq) = (["N"_2"O"_4])/(["NO"_2]^color(red)(2))`

This means that you have

`["N"_ 2"O"_ 4] = K_(eq) * ["NO"_ 2]^color(red)(2)`

Plug in your values to find--since you didn't get units for `K_(eq)`, 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.