Question

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"`?

Answer 1

`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.