An equilibrium mixture in a vessel of capacity #"100 L"# contains #1# mole of #"N"_2#, #2# moles of #"O"_2#, and #3# of #"NO"#. No.of moles of #"O"_2# to be added so that, at new equilibrium, the concentration of #"NO"# is found to be #"0.04 mol/L"#?
1 Answer
Explanation:
The first thing that you need to do here is to calculate the equilibrium constant of the reaction at this particular temperature.
You know that
#"O"_ (2(g)) + "N"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))#
For this reaction, the equilibrium constant,
#K_c = (["NO"]^2)/(["O"_2] * ["N"_2])#
The equilibrium concentrations of the three chemical species will be
#["O"_2] = "2 moles"/"100 L" = "0.02 M"#
#["N"_2] = "1 mole"/"100 L" = "0.01 M"#
#["NO"] = "3 moles"/"100 L" = "0.03 M"#
This means that you have--I'll leave the expression of the equilibrium constant without added units!
#K_c = (0.03)^2/(0.02 * 0.01)#
#K_c= (3^2 * color(red)(cancel(color(black)(1/100^2))))/(2^2 * color(red)(cancel(color(black)(1/100))) * 1 * color(red)(cancel(color(black)(1/100))))#
#K_c = 9/2#
Now, you know that after some oxygen gas is added to the reaction vessel, the equilibrium concentration of nitrogen oxide is equal to
#["NO"]_"new" = "0.04 M"#
The balanced chemical equation tells you that in order for the reaction to produce
This means that in order for the concentration of nitrogen oxide to increase by
#"0.04 M " - " 0.03 M" = "0.01 M"#
the concentrations of nitrogen gas and of oxygen gas must decrease by
#["N"_ 2]_ "new" = ["N"_ 2] - 1/2 * "0.01 M"#
#["N"_ 2]_ "new" = "0.01 M" - 1/2 * "0.01 M"#
#["N"_ 2]_ "new" = "0.005 M"#
Next, use the expression of the equilibrium constant to find the new equilibrium concentration of oxygen gas
#["O"_ 2]_ "new" = (["NO"]_ "new"^2)/(K_c * ["N"_ 2]_ "new")#
#["O"_ 2]_ "new" = (0.04)^2/(9/2 * 0.005) = 16/225#
This means that, when the new equilibrium is established, you have
#["O"_ 2]_ "new" = 16/225 quad "M"#
Use the fact that in order for the new equilibrium to be established, the reaction consumed
#["O"_ 2]_"increased" = 16/225 quad "M" + 1/2 * "0.01 M"#
#["O"_ 2]_ "increased" = 137/1800 quad "M"#
This means that the concentration of oxygen gas increased by
#Delta_( ["O"_ 2]) = 137/1800 quad "M" - "0.02 M"#
#Delta_ (["O"_ 2]) = 101/1800 quad "M"#
Finally, to find the number of moles of oxygen gas added to the reaction vessel, use the volume of the vessel
#100 color(red)(cancel(color(black)("L"))) * (101/1800 quad "moles O"_2)/(1color(red)(cancel(color(black)("L")))) = 101/18 quad "moles O"_2#
You should round this off to one significant figure to get
#"moles of O"_2 ~~ "6 moles"#
but I'll leave the answer in fraction form.