An equilibrium mixture in a vessel of capacity $100 L$ $"100 L"$ contains $1$ $1$ mole of $N_{2}$ $"N"_2$ , $2$ $2$ moles of $O_{2}$ $"O"_2$ , and $3$ $3$ of $NO$ $"NO"$ . No.of moles of $O_{2}$ $"O"_2$ to be added so that, at new equilibrium, the concentration of $NO$ $"NO"$ is found to be $0.04 mol/L$ $"0.04 mol/L"$ ?

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An equilibrium mixture in a vessel of capacity $100 L$ $"100 L"$ contains $1$ $1$ mole of $N_{2}$ $"N"_2$ , $2$ $2$ moles of $O_{2}$ $"O"_2$ , and $3$ $3$ of $NO$ $"NO"$ . No.of moles of $O_{2}$ $"O"_2$ to be added so that, at new equilibrium, the concentration of $NO$ $"NO"$ is found to be $0.04 mol/L$ $"0.04 mol/L"$ ?

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Answer 1 · 2018-01-14T15:36:58

$\frac{101}{18}$ $101/18$ ${moles O}_{2}$ $"moles O"_2$

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

$O_{2 (g)} + N_{2 (g)} ⇌ 2 {NO}_{2 (g)}$ $"O"_ (2(g)) + "N"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))$

For this reaction, the equilibrium constant, $K_{c}$ $K_c$ , is equal to

$K_{c} = \frac{{[NO]}^{2}}{[O_{2}] \cdot [N_{2}]}$ $K_c = (["NO"]^2)/(["O"_2] * ["N"_2])$

The equilibrium concentrations of the three chemical species will be

$[O_{2}] = \frac{2 moles}{100 L} = 0.02 M$ $["O"_2] = "2 moles"/"100 L" = "0.02 M"$

$[N_{2}] = \frac{1 mole}{100 L} = 0.01 M$ $["N"_2] = "1 mole"/"100 L" = "0.01 M"$

$[NO] = \frac{3 moles}{100 L} = 0.03 M$ $["NO"] = "3 moles"/"100 L" = "0.03 M"$

This means that you have--I'll leave the expression of the equilibrium constant without added units!

$K_{c} = \frac{{(0.03)}^{2}}{0.02 \cdot 0.01}$ $K_c = (0.03)^2/(0.02 * 0.01)$

$K_c= (3^2 * color(red)(cancel(color(black)(1/100^2))))/(2^2 * color(red)(cancel(color(black)(1/100))) * 1 * color(red)(cancel(color(black)(1/100))))$

$K_c = 9/2$

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.

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