# Question #e0529

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The chemical equation that describes this equilibrium looks like this

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

The **equilibrium constant** is calculated by using the equation

#K_c = (["NO"_2]^color(red)(2))/(["N"_2"O"_4])#

Now, you know that at

#K_c = 5.88 * 10^(-3)#

The fact that the equilibrium constant is *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 **moles** of nitrogen dioxide.

This means that if you take **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

#color(red)(2)x# #"M"# of nitrogen dioxide, it mustconsume#x# #"M"# of dinitrogen tetroxide.

Plug this back into the expression of the equilibrium constant

#K_c = (color(red)(2)x)^color(red)(2)/(0.03991 - x)#

This is equivalent to

#5.88 * 10^(-3) = (4x^2)/(0.03991 - x)#

Rearrange to quadratic equation form

#4x^2 + 5.88 * 10^(-3) * x - 0.03991 * 5.88 * 10^(-3) = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since you're looking for *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.