# At 25^@ "C", for the reaction "N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g), the equilibrium constant is K_c = 5.85xx10^(-3). "20.0 g" of "N"_2"O"_4 were added to a "5.00-L" container. What is the molar concentration of "NO"_2 at equilibrium?

## N_2O_4(g) ⇌ 2NO_2(g)

Jan 2, 2017

I got $\text{0.0146 M}$.

Since you know the chemical formula for dinitrogen tetroxide, you can calculate its molar mass, which allows you to get the initial molar concentration.

If we define the molar mass of substance $\boldsymbol{i}$ as ${M}_{m , i}$, then:

${M}_{m , {N}_{2} {O}_{4}} = 2 {M}_{m , N} + 4 {M}_{m , O}$

$= 2 \left(\text{14.007 g/mol") + 4("15.999 g/mol}\right)$

$=$ $\text{92.01 g/mol}$

Therefore, the $\text{mol}$s of this substance added to the container are:

color(green)(n_(N_2O_4)) = 20.0 cancel("g N"_2"O"_4(g)) xx "1 mol"/(92.01 cancel("g N"_2"O"_4(g)))

$=$ $\textcolor{g r e e n}{\text{0.217 mols}}$

and so, you can find the concentration that it was initially at, by assuming that as a gas, it fills the entire container (i.e. use the volume of the container as your volume):

color(green)(["N"_2"O"_4(g)]_0) = ("0.217 mols")/("5.00 L") = color(green)("0.0435 M")

At this point, you have enough information to do the remainder of the problem.

The balanced reaction was:

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

so if $x$ concentration of ${\text{N"_2"O}}_{4} \left(g\right)$ was stoichiometrically transformed into $2 x$ concentration of ${\text{NO}}_{2} \left(g\right)$ (not $x$!), then

${\left[{\text{N"_2"O}}_{4} \left(g\right)\right]}_{e q} = 0.0435 - x$ $\text{M}$,

${\left[{\text{NO}}_{2} \left(g\right)\right]}_{e q} = 2 x$ $\text{M}$,

and:

${K}_{c} = 5.85 \times {10}^{- 3} = \left(\left[{\text{NO"_2(g)]^2)/(["N"_2"O}}_{4} \left(g\right)\right]\right)$

$= {\left(2 x\right)}^{2} / \left(0.0435 - x\right) = \frac{4 {x}^{2}}{0.0435 - x}$

Since ${K}_{c} > {10}^{- 5}$, it is probably not a good idea to make a small $x$ approximation, but we will give that a shot and compare to the true answer.

WITH THE SMALL X APPROXIMATION

$5.85 \times {10}^{- 3} \cdot 0.0435 \approx 4 {x}^{2}$

$\implies \textcolor{g r e e n}{x} \approx | \sqrt{\frac{1}{4} \left(5.85 \times {10}^{- 3} \cdot 0.0435\right)} |$

$=$ $\textcolor{red}{\text{0.00797 M}}$

WITHOUT THE SMALL X APPROXIMATION

$5.85 \times {10}^{- 3} \cdot 0.0435 - 5.85 \times {10}^{- 3} x - 4 {x}^{2} = 0$

which we solve using the quadratic formula to get:

$\textcolor{g r e e n}{x} = \frac{- \left(- 5.85 \times {10}^{- 3}\right) \pm \sqrt{{\left(- 5.85 \times {10}^{- 3}\right)}^{2} - 4 \left(- 4\right) \left(5.85 \times {10}^{- 3} \cdot 0.0435\right)}}{2 \left(- 4\right)}$

$= \left[\ldots\right]$

which, if you take the physically reasonable answer, gives you

$=$ $\textcolor{g r e e n}{\text{0.00728 M}}$

which is 9.59% error (more than an acceptable 5% error), and so, that is why you should not make a small $x$ approximation when ${K}_{c} > {10}^{- 5}$ (same with ${K}_{p}$ and ${K}_{s p}$).

So, we take this more accurate answer, and plug it back into the original expression for ${\left[{\text{NO}}_{2} \left(g\right)\right]}_{e q}$ to get:

$\textcolor{b l u e}{{\left[{\text{NO}}_{2} \left(g\right)\right]}_{e q}} = 2 x$

$= 2 \left(\text{0.00728 M}\right)$

$=$ $\textcolor{b l u e}{\text{0.0146 M}}$