Question #9bbc5
1 Answer
Here's what I got.
Explanation:
Start by writing the balanced chemical equation for this equilibrium reaction
#color(red)(2)"NO"_ (2(g)) rightleftharpoons color(blue)(2)"NO"_ ((g)) + "O"_(2(g))#
The first thing to notice here is that the equilibrium constant for this reaction
#K_c = 1.2 * 10^(-5)#
is smaller than
In other words, you can expect the equilibrium concentrations of the two reactants, nitric oxide,
Use the number of moles of nitrogen dioxide and the volume of the reaction vessel to determine the initial concentration of the reactant
#["NO"_2] = "0.50 moles"/"2.0 L" = "0.25 M"#
Use an ICE table to help you determine the equilibrium concentrations of the three chemical species
#" "color(red)(2)"NO"_ (2(g)) " "rightleftharpoons" " color(blue)(2)"NO"_ ((g)) " "+" " "O"_(2(g))#
By definition, the equilibrium constant for this reaction will look like this
#K_c = (["O"_2] * ["NO"]^color(blue)(2))/(["NO"_2]^color(red)(2))#
In this case, you will have
#K_c =(x * (color(blue)(2)x)^color(blue)(2))/((0.25 - color(red)(2)x)^color(red)(2)#
#K_c = (4x^3)/(0.25 - 2x)^2#
Now, because
#0.25 - 2x ~~ 0.25#
This means that you'll have
#K_c = (4x^3)/0.25^2#
This will get you
#x = root(3)( (K_c * 0.25^2)/4) = root(3)((1.2 * 10^(-5) * 0.0625)/4)#
#x = 0.005724#
The equilibrium concentrations of the three species will thus be
#["NO"_2] = "0.25 M" - color(red)(2) * "0.005724 M" = "0.2386 M"#
#["NO"] = color(blue)(2) * "0.005724 M" = "0.01145 M"#
#["O"_2] = "0.005724 M"#
Rounded to two sig figs, the answers will be
#["NO"_2] = color(green)(|bar(ul(color(white)(a/a)"0.24 M"color(white)(a/a)|)))#
#["NO"] = color(green)(|bar(ul(color(white)(a/a)"0.011 M"color(white)(a/a)|)))#
#["O"_2] = color(green)(|bar(ul(color(white)(a/a)"0.0057 M"color(white)(a/a)|)))#
As predicted, the equilibrium concentrations of the products are significantly smaller than the equilibrium concentration of the reactant.