Question

Density of equilibrium mixture of ${N_2}{O_4}$ and $N{O_2}$ at $1atm$ and $384K$ is $1.84gd{m^{ - 3}}$ . Equilibrium constant of the following reaction is:
${N_2}{O_4}\underset {} \leftrightarrows 2N{O_2}$.
A: $1.98atm$
B: $2.09atm$
C: $2.36atm$
D: $1.48atm$

Answer 1

Equilibrium constant is the value of reaction quotient at equilibrium. This constant depends on the temperature of the reaction. Partial pressure is the pressure which is exerted by a gas if it would be the only gas present in the system. Sum of partial pressure of all the gases present in a mixture is equal to the total pressure.

Formula used: Equilibrium constant ${K_p} = \dfrac{{{{\left( {{p_{N{O_2}}}} \right)}^2}}}{{{p_{_{{N_2}{O_4}}}}}}$
$PM = dRT$
$R = 0.082d{m^3}.atm.{K^{ - 1}}.mo{l^{ - 1}}$
$x = \dfrac{{D - d}}{{\left( {n - 1} \right)d}}$
Mole fraction $= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
Partial pressure $=$total pressure$\times$mole fraction of particular gas
Vapour density $= \dfrac{1}{2} \times$mass
Where, $P$ is total pressure, $M$ is the mass of mixture, $d$ is density, $R$ is gas constant, $T$ is temperature, $x$ is number of moles of reactant consumed, $D$ initial vapour density, $d$ vapour density at equilibrium, $n$ is number of moles of gaseous product for one mole of reactant ${p_{N{O_2}}}$ is partial pressure of $N{O_2}$ and ${p_{{N_2}{O_4}}}$ is partial pressure of ${N_2}{O_4}$.

Complete step by step answer:
Vapour density of a particular gas or vapour is defined as the mass of gas or vapour as compared to the equal volume of hydrogen. Partial pressure is the pressure which is exerted by a gas if it would be the only gas present in the system. Sum of partial pressure of all the gases present in a mixture is equal to the total pressure.
In this question we have to calculate equilibrium constant (in terms of pressure) of the following reaction at equilibrium:
${N_2}{O_4} \underset {} \leftrightarrows 2N{O_2}$
Equilibrium constant is the value of reaction quotient at equilibrium. This constant depends on the temperature of the reaction. To calculate equilibrium constant first we have to calculate the mass of the mixture. Mass of mixture can be calculated by using the following formula,
$PM = dRT$
Pressure $P$ is $1atm$ (given)
Density is $d$ is $1.84gd{m^{ - 3}}$ (given)
Temperature $T$ is $384K$ (given)
Gas constant $R = 0.082d{m^3}.atm.{K^{ - 1}}.mo{l^{ - 1}}$
Substituting these values in the formula we can calculate mass of the mixture.
$1 \times M = 1.84 \times 0.082 \times 384$
Solving this we get,
$M = 58g$
So, the mass of the mixture is $58g$. From this we can calculate vapour density at equilibrium using the following formula:
Vapour density $= \dfrac{1}{2} \times$mass
Mass is $58g$. So, vapour density at equilibrium is:
Vapour density at equilibrium $d = \dfrac{1}{2} \times 58 = 29$
Mass of ${N_2}{O_4}$ is $92g$ (calculated by adding individual mass of all the atoms present in this compound). Initially only ${N_2}{O_4}$ was present. So, the initial vapour density will be calculated by using mass of ${N_2}{O_4}$. Therefore,
Initial vapour density$D = \dfrac{1}{2} \times 92 = 46$
Let us suppose initially there was one mole of ${N_2}{O_4}$ and at the equilibrium $x$ moles of ${N_2}{O_4}$ are consumed. Number of moles consumed can be calculated by the formula,
$x = \dfrac{{D - d}}{{\left( {n - 1} \right)d}}$
Number of moles of gaseous products is two (from reaction)
Substituting the values calculated above we get,
$x = \dfrac{{46 - 29}}{{\left( {2 - 1} \right)29}}$
Solving this we get,
$x = 0.586$
So number of moles that are consumed is $0.586$
Since, there was only one mole of ${N_2}{O_4}$ initially and at the equilibrium $x$ moles of ${N_2}{O_4}$ are consumed. According to the equilibrium reaction,
${N_2}{O_4}\underset {} \leftrightarrows 2N{O_2}$
If $x$ moles of ${N_2}{O_4}$ are consumed then $2x$ moles of $N{O_2}$ are formed. So, at equilibrium, concentration of ${N_2}{O_4}$ is $1 - x$ ($x$ moles of ${N_2}{O_4}$ are consumed) and concentration of $N{O_2}$ will be $2x$. At equilibrium total number of moles will be,
Total moles$= 1 - x + 2x = 1 + x$
From this we can calculate mole fraction of ${N_2}{O_4}$ and $N{O_2}$ at equilibrium using the following formula,
Mole fraction$= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
Moles of ${N_2}{O_4}$ is $1 - x$ and total moles are calculated above. So, mole fraction of ${N_2}{O_4}$ is,
Mole fraction of ${N_2}{O_4}$ $= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
Mole fraction of ${N_2}{O_4}$ $= \dfrac{{1 - x}}{{1 + x}}$
Moles of $N{O_2}$ is $2x$ and total moles are calculated above. So, mole fraction of $N{O_2}$ is,
Mole fraction of $N{O_2}$ $= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
Mole fraction of $N{O_2}$ $= \dfrac{{2x}}{{1 + x}}$
From this information we can calculate partial pressure of gases using the formula,
Partial pressure$=$total pressure$\times$mole fraction of particular gas
For $N{O_2}$,
Partial pressure ${p_{N{O_2}}}$ $=$ $1 \times \dfrac{{2x}}{{1 + x}}$ (given pressure is $1atm$)
For ${N_2}{O_4}$,
Partial pressure ${p_{{N_2}{O_4}}}$ $=$ $1 \times \dfrac{{1 - x}}{{1 + x}}$ (given pressure is $1atm$)
With the help of this partial pressure we can calculate equilibrium constant using the formula,
Equilibrium constant ${K_p} = \dfrac{{{{\left( {{p_{N{O_2}}}} \right)}^2}}}{{{p_{_{{N_2}{O_4}}}}}}$
Substituting the values we get,
Equilibrium constant ${K_p} = \dfrac{{{{\left( {\dfrac{{2x}}{{1 + x}}} \right)}^2}}}{{\left( {\dfrac{{1 - x}}{{1 + x}}} \right)}}$
Simplifying this equation,
Equilibrium constant ${K_p} = \dfrac{{4{x^2}}}{{1 - {x^2}}}$
Substituting the values of $x$ calculated above, that is $0.586$,
Equilibrium constant ${K_p} = \dfrac{{4{{\left( {0.586} \right)}^2}}}{{1 - {{\left( {0.586} \right)}^2}}}$
Solving this we get,
Equilibrium constant ${K_p} = 2.09$

So, the equilibrium constant of this reaction is $0.29atm$ and the correct answer is option B.

Note:
Equilibrium constant in terms of concentration can be calculated by dividing the concentration of products each raised to the power number of moles produced with concentration of reactants consumed each raised to the power number of moles consumed.