Question

Consider the following equilibrium in a closed container ${N_2}{O_{4\left( g \right)}} \rightleftharpoons 2N{O_{2\left( g \right)}}$. At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements regarding the equilibrium constant $\left( {{K_p}} \right)$ and degree of dissociation $\left( \alpha \right)$:-
Options-
A.Neither $\left( {{K_p}} \right)$ nor change $\left( \alpha \right)$
B.Both $\left( {{K_p}} \right)$ and $\left( \alpha \right)$ changes
C.$\left( {{K_p}} \right)$ changes, but $\left( \alpha \right)$ does not changes
D.$\left( {{K_p}} \right)$ does not change, but $\left( \alpha \right)$ changes

Answer 1

Hint: Van’t Hoff principle is used to calculate degree of dissociation which is defined as the ratio of total number of moles of substance dissociated to total number of moles of substance taken initially. Equilibrium constant is defined as ratio of rate of forward reaction and ratio of backward reaction.

Complete answer:
Equilibrium constant of chemical reaction is independent of some factors like initial concentration of reactant, or presence or absence of catalyst. But the value of Equilibrium constant is dependent on the nature of initial reactant and final product, temperature at which reaction is carried out as well as pressure of reaction.
When reaction proceeds it generates two molecules of nitrogen oxide based on its dissociation constant $\left( \alpha \right)$. The chemical reaction in terms of $\left( \alpha \right)$ is expressed as-
${N_2}{O_{4\left( g \right)}} \rightleftharpoons 2N{O_{2\left( g \right)}}$
$\left( {1 - \alpha } \right)$ $\left( {2\alpha } \right)$
Since, the temperature of the reaction does not change, the value of Equilibrium constant $\left( {{K_p}} \right)$ does not change. When volume of reaction decreased by half of its initial value pressure of reaction mixture doubles as a result of which value of Equilibrium constant $\left( {{K_p}} \right)$ increases theoretically but practically value of $\left( {{K_p}} \right)$ does not changes. Therefore, in order to maintain the value at constant other parameters changes. Due to change in volume of reaction concentration of the reactant changes and due to this value of dissociation constant $\left( \alpha \right)$ also changes because it depends on the concentration of the reactant. In order to maintain the value of Equilibrium constant $\left( {{K_p}} \right)$ at constant value of dissociation constant $\left( \alpha \right)$changes accordingly.
$ \Rightarrow $$\left( {{K_p}} \right)$ does not change, but $\left( \alpha \right)$ changes.

Therefore, option $\left( D \right)$ is the correct option.

Note:
There is no standard unit of equilibrium constant because it is a ratio of two similar properties. Any pure substance used during the chemical reaction is not expressed while writing the equilibrium constant of the reaction. Le-Chatelier’s principle is used to express the relation of pressure and Equilibrium constant.