# 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"?

Jan 14, 2018

$\frac{101}{18}$ ${\text{moles O}}_{2}$

#### 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

${\text{O"_ (2(g)) + "N"_ (2(g)) rightleftharpoons 2"NO}}_{2 \left(g\right)}$

For this reaction, the equilibrium constant, ${K}_{c}$, is equal to

${K}_{c} = \left(\left[{\text{NO"]^2)/(["O"_2] * ["N}}_{2}\right]\right)$

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!

${K}_{c} = {\left(0.03\right)}^{2} / \left(0.02 \cdot 0.01\right)$

${K}_{c} = \frac{{3}^{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{100} ^ 2}}}}{{2}^{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{100}}}} \cdot 1 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{100}}}}}$

${K}_{c} = \frac{9}{2}$

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 $\frac{1}{2}$ moles of oxygen gas and $\frac{1}{2}$ moles of nitrogen gas.

This means that in order for the concentration of nitrogen oxide to increase by

$\text{0.04 M " - " 0.03 M" = "0.01 M}$

the concentrations of nitrogen gas and of oxygen gas must decrease by $\frac{1}{2} \cdot \text{0.01 M}$. 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

$\left[\text{O"_ 2]_ "new" = (["NO"]_ "new"^2)/(K_c * ["N"_ 2]_ "new}\right)$

["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 $\frac{1}{2} \cdot \text{0.01 M}$ 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

$\text{moles of O"_2 ~~ "6 moles}$

but I'll leave the answer in fraction form.