Question

An equilibrium mixture in a vessel of capacity 100 liter can $1\,mol\,{N_2}$, $2\,mol\,{O_2}$ and $3\,mol\,NO$. Number of moles of ${O_2}$ to be so that the new equilibrium the concentration of $NO$is found to be $0.04\,mol/lit$ is:
A. $101/18$
B. $101/9$
C. $202/9$
D.None of these

Answer 1

We can calculate the moles of oxygen gas using the molarity and volume. Before that, we have to write the rate of the reaction and then calculate the equilibrium constant of the reaction. The equilibrium constant is calculated using the concentration rise to power of the coefficient in the balanced chemical equation.

Complete answer: or Complete step by step answer:
Let us now calculate the equilibrium constant for the reaction at specific temperature. We can write the balanced chemical reaction as,
${O_2}\left( g \right) + {N_2}\left( g \right)\xrightarrow{{}}2N{O_2}\left( g \right)$
We can write the expression for equilibrium constant ${K_c}$ as,
${K_c} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{O_2}} \right]\left[ {{N_2}} \right]}}$
The moles of oxygen is $2\,mol$.
The moles of nitrogen is $1\,mol$.
The moles of nitrogen oxide is $3\,mol$.
The volume is $100\,L$.
The formula to calculate the equilibrium concentration,
${\text{Concentration = }}\dfrac{{{\text{Mass}}\left( {{\text{in}}\,{\text{mol}}} \right)}}{{{\text{Volume}}\left( {{\text{in}}\,{\text{L}}} \right)}}$
We can calculate the equilibrium concentration for the three chemical species using the moles and volume.
${\text{Concentration = }}\dfrac{{{\text{Mass}}\left( {{\text{in}}\,{\text{mol}}} \right)}}{{{\text{Volume}}\left( {{\text{in}}\,{\text{L}}} \right)}} \\\ \left[ {{O_2}} \right] = \dfrac{{2\,mol}}{{100\,L}} = 0.02\,M \\\ \left[ {{N_2}} \right] = \dfrac{{1\,mol}}{{100\,L}} = 0.01\,M \\\ \left[ {NO} \right] = \dfrac{{3\,mol}}{{100\,L}} = 0.03\,M \\\$
The concentration of oxygen gas is $0.02\,M$.
The concentration of nitrogen gas is $0.01M$.
The concentration of nitrogen oxide is $0.03\,M$.
From the calculated concentration, let us now calculate the equilibrium concentration ${K_c}$.
We have calculate the equilibrium constant by substituting the values of concentrations of $N{O_2}$, ${O_2}$ and ${N_2}$
${K_c} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{O_2}} \right]\left[ {{N_2}} \right]}} \\\ {K_c} = \dfrac{{{{\left( {0.03} \right)}^2}}}{{\left( {0.02} \right)\left( {0.01} \right)}} \\\ {K_c} = \dfrac{{{3^2} \times \dfrac{1}{{{{100}^2}}}}}{{2 \times \dfrac{1}{{100}} \times 1 \times \dfrac{1}{{100}}}} \\\ {K_c} = \dfrac{9}{2} \\\$
The value of equilibrium concentration is $\dfrac{9}{2}$.
After some amount of oxygen gas is added to the vessel, the concentration of nitrogen oxide is equal to $0.04\,M$.
We know the balanced chemical reaction says that in order of the reaction to yield $1$ mole of nitrogen oxide, it should consume $\dfrac{1}{2}\,mol$ of oxygen gas and $\dfrac{1}{2}\,mol$ of nitrogen gas. This means that in order for the concentration of nitrogen oxide to increase by $0.04\,M - 0.03\,M = 0.01\,M$.
We must decrease the concentration of oxygen and nitrogen gas by $\dfrac{1}{2} \times 0.01\,M$. The value of new equilibrium concentration of nitrogen gas has to be calculated.
${\left[ {{N_2}} \right]_{new}} = \left[ {{N_2}} \right] - \dfrac{1}{2} \times 0.01\,M \\\ {\left[ {{N_2}} \right]_{new}} = 0.01\,M - \dfrac{1}{2} \times 0.01\,M \\\ {\left[ {{N_2}} \right]_{new}} = 0.005\,M \\\$
The new equilibrium concentration of nitrogen gas is $0.005\,M$.
We can find the new equilibrium constant of oxygen gas using the equilibrium constant.
${\left[ {{O_2}} \right]_{new}} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{{K_c} \times {{\left[ {{N_2}} \right]}_{new}}}} \\\ {\left[ {{O_2}} \right]_{new}} = \dfrac{{{{\left( {0.04} \right)}^2}}}{{\dfrac{9}{2} \times 0.005}} \\\ {\left[ {{O_2}} \right]_{new}} = \dfrac{{16}}{{225}}\,M \\\$
We have calculated the new equilibrium concentration of oxygen as $\dfrac{{16}}{{225}}\,M$.
The reaction consumed $\dfrac{1}{2} \times 0.01\,M$ of oxygen gas to find the concentration of oxygen gas after the number of moles of this reactant increased.
We can calculate the concentration of new oxygen gas as,
${\left[ {{O_2}} \right]_{increased}} = \dfrac{{16}}{{225}}\,M + \dfrac{1}{2} \times 0.01\,M \\\ {\left[ {{O_2}} \right]_{increased}} = \dfrac{{137}}{{1800}}\,M \\\$
The equilibrium concentration of new oxygen gas is $\dfrac{{137}}{{1800}}\,M$.
The amount of concentration of oxygen gas is increased by,
$\Delta \left[ {{O_2}} \right] = {\text{New}}\,{\text{concentration - Original}}\,{\text{concentration}} \\\ \Delta \left[ {{O_2}} \right] = \dfrac{{137}}{{1800}}M - 0.02\,M \\\ \Delta \left[ {{O_2}} \right] = \,\dfrac{{101}}{{1800}}M \\\$
The amount of concentration of oxygen gas is $\dfrac{{101}}{{1800}}\,M$.
We can calculate the number of moles of oxygen gas added to the vessel using the concentration and volume.
${\text{Number}}\,{\text{of}}\,{\text{moles = Concentration \times Volume}} \\\ Number\,of\,moles = \dfrac{{\dfrac{{101}}{{1800}}\,moles\,{O_2}}}{{1\,L}} \times 100\,L \\\ Number\,of\,moles = \dfrac{{101}}{{18}}\,moles\,{O_2} \\\$
The number of moles of oxygen is $\dfrac{{101}}{{18}}\,mol$.

$\therefore$Option (A) is correct.

Note:
We can express the equilibrium constant in terms of pressure as ${K_p}$. We can convert ${K_c}$ to ${K_p}$ using the equation,
${K_p} = {K_c}{\left( {RT} \right)^{\Delta {n_{gas}}}}$
Here, R=Gas constant $\left( {8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)$
T=Temperature (in Kelvin scale)
${{\Delta }}{{\text{n}}_{{\text{gas}}}}{\text{ = Mass}}\,{\text{of}}\,{\text{gas}}\left( {{\text{product}}} \right){\text{ - Mass}}\,{\text{of}}\,{\text{gas}}\,\left( {{\text{reactant}}} \right)$
The value of equilibrium constant is not affected by the rate of catalyst in forward and backward reaction.
We have different values of equilibrium constant for the same reversible reaction.
The reciprocal of original equilibrium is the equilibrium constant for the reverse equilibrium.
We can predict the extent of the reaction using the equilibrium constant. The magnitude of the equilibrium constant tells about the relative amounts of the reactants and the products.
We have to remember that if the value of equilibrium constant is large, forward reaction is favored and a reverse reaction is favored when the value of equilibrium constant is very low.