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
Jan 14, 2018

Answer:

#101/18# #"moles O"_2#

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#, is equal to

#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 #1# mole of nitrogen oxide, it must consume #1/2# moles of oxygen gas and #1/2# moles of nitrogen gas.

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 #1/2 * "0.01 M"#. Since you didn't add any nitrogen gas to the reaction vessel, you can say that the new equilibrium concentration of nitrogen gas will be

#["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 #1/2 * "0.01 M"# of oxygen gas to find the concentration of oxygen gas after you increased the number of moles of this reactant,

#["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.