# Question e0529

Feb 16, 2018

Here's what I got.

#### Explanation:

The chemical equation that describes this equilibrium looks like this

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

The equilibrium constant is calculated by using the equation

${K}_{c} = \left(\left[{\text{NO"_2]^color(red)(2))/(["N"_2"O}}_{4}\right]\right)$

Now, you know that at ${25}^{\circ} \text{C}$, you have

${K}_{c} = 5.88 \cdot {10}^{- 3}$

The fact that the equilibrium constant is $< 1$ tells you that, at this temperature, the reverse reaction is favored, meaning that the equilibrium lies to the left.

Consequently, you can expect the equilibrium concentration of dinitrogen tetroxide to be higher than the equilibrium concentration of nitrogen dioxide.

So, use the molar mass of dinitrogen tetroxide to calculate the number of moles present in your sample.

15.6 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"O"_4)/(92.011 color(red)(cancel(color(black)("g")))) = "0.16954 moles N"_2"O"_4

To find the initial concentration of dinitrogen tetroxide, divide the number of moles by the volume of the flask.

["N"_2"O"_4] = "0.16954 moles"/("5.00 L") = "0.03391 M"

Now, you know that every mole of dinitrogen tetroxide that reacts to produce nitrogen dioxide will produce $\textcolor{red}{2}$ moles of nitrogen dioxide.

This means that if you take $x$ $\text{M}$ to be the concentration of dinitrogen tetroxide that reacts, you will have

["NO"_2] = color(red)(2) * x quad "M"

["N"_2"O"_4] = (0.03991 - x) quad "M"

In order for the reaction to produce $\textcolor{red}{2} x$ $\text{M}$ of nitrogen dioxide, it must consume $x$ $\text{M}$ of dinitrogen tetroxide.

Plug this back into the expression of the equilibrium constant

${K}_{c} = {\left(\textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(0.03991 - x\right)$

This is equivalent to

$5.88 \cdot {10}^{- 3} = \frac{4 {x}^{2}}{0.03991 - x}$

$4 {x}^{2} + 5.88 \cdot {10}^{- 3} \cdot x - 0.03991 \cdot 5.88 \cdot {10}^{- 3} = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since you're looking for $x$ to represent concentration, you can discard the negative solution and say that

$x = 0.0063634$

This means that, at equilibrium, the flask will contain

["N"_2"O"_4] = "0.03391 M "- " 0.0063634 M" = color(darkgreen)(ul(color(black)("0.0275 M")))

["NO"_2] = color(red)(2) * "0.0063634 M" = color(darkgreen)(ul(color(black)("0.0127 M")))#

I'll leave both values rounded to three sig figs.

Notice that, as predicted, the equilibrium concentration of dinitrogen pentoxide is higher than the equilibrium concentration of nitrogen dioxide.