Question

The value of ${K_p}$ for the equilibrium reaction ${N_2}{O_4}(g)\underset {} \leftrightarrows 2N{O_2}(g)$ is $2$. The percentage dissociation of ${N_2}{O_4}(g)$ at a pressure of $0.5\,atm$ is
A.$71$
B.$50$
C.$88$
D.$25$

Answer 1

Hint: While dealing with the degree of dissociation $\alpha $, ${K_p}$ and number of moles, you can assume the initial number of moles of reactants to be one. This will lead to an easier calculation. You should take care of the stoichiometry of the reaction while dealing with the degree of dissociation $\alpha $.

Formula used:
${K_p} = {K_n}{\left( {\dfrac{{{P_T}}}{{{n_T}}}} \right)^{\Delta {n_g}}}$
Where ${P_T}$ is the total pressure of the system at equilibrium
${n_T}$ is the total number of moles of the system at equilibrium
$\Delta {n_g}$ is the difference in number of gaseous species in products and reactants
${K_p}$ is the equilibrium constant when partial pressure is considered.
${K_n}$ is the equilibrium constant when number of moles are considered.

Complete step-by-step answer:Dissociation of ${N_2}{O_4}$ proceeds according to the equation:
${N_2}{O_4}(g)\underset {} \leftrightarrows 2N{O_2}(g)$
$1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$ Initial number of moles
$1 - \alpha $ $2\alpha $ Number of moles at equilibrium
Here, we will be assuming initial moles of the reactant i.e., ${N_2}{O_4}$to be one. Since zero moles of product will be present in the beginning. It's written as zero. Let us assume the degree of dissociation to be $\alpha $. Number of moles at equilibrium will be $1 - \alpha $and $2\alpha $for reactants and products respectively.
Total number of moles at equilibrium ${n_T}$ $ = 1 - \alpha + 2\alpha $
 Total number of moles at equilibrium $ = 1 + \alpha $
Given value of total pressure at equilibrium ${P_T}$ $ = 0.5\,atm$
Applying law of chemical equilibrium,
${K_p} = \dfrac{{{{\left( {{n_{N{O_2}}}} \right)}^2}}}{{{n_{{N_2}{O_4}}}}} \times \dfrac{{{P_T}}}{{{n_T}}}$
Substituting the values, we will get
Putting the equilibrium number of moles of reactants and products,
$
  2 = \dfrac{{{{\left( {2\alpha } \right)}^2}}}{{1 - \alpha }} \times \dfrac{{0.5}}{{1 + \alpha }} \\
 \Rightarrow 2 = \dfrac{{4{\alpha ^2}}}{{1 - {\alpha ^2}}} \times \dfrac{1}{2} \\
  \Rightarrow 4\left( {1 - {\alpha ^2}} \right) = 4{\alpha ^2} \\
   \Rightarrow 1 = 2{\alpha ^2} \\
   \Rightarrow 0.5 = {\alpha ^2} \\
 $
$\therefore 0.71 = \alpha $
The percentage dissociation will be given by: $
 \Rightarrow Percentage dissociation = \alpha \times 100 \\
 \Rightarrow Percentage dissociation = 0.71 \times 100 \\
 \Rightarrow Percentage dissociation = 71 \\
 $

Hence, option A is the correct answer.

Note:We should remember the relation between different equilibrium constants while doing these questions. These relations help us in solving these questions with less difficulty. Always remember the expressions for equilibrium constants depend on the equilibrium values rather than initial values. The expression for ${K_p}$ will depend on the equilibrium partial pressure of all the species rather than depending on the initial values.