Question #e0529
1 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
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
This means that if you take
#["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 must consume#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
#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.