Question #d78b6

1 Answer
Jul 19, 2017

Answer:

#["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.