Question

The degree of dissociation ‘\[\alpha \]’ of the reaction:
\[{N_2}{O_4} \rightleftharpoons 2N{O_{2(g)}}\]
Can be related to \[{K_p}\] as:
A.\[\alpha = \dfrac{{\dfrac{{{K_p}}}{P}}}{{1 + \dfrac{{{K_p}}}{P}}}\]
B.\[\alpha = \dfrac{{{K_p}}}{{4 + {K_p}}}\]
C.\[\alpha = \dfrac{1}{{\sqrt {\dfrac{{{K_p}/P}}{{4 + {K_p}/P}}} }}\]
D.\[\alpha = \sqrt {\dfrac{{{K_p}}}{{4 + {K_p}}}} \]

Answer 1

Hint: The dissociation degree is the fraction of original solute molecules that have dissociated. It is usually indicated by the Greek symbol α. More accurately, degree of dissociation refers to the amount of solute dissociated into ions or radicals per mole.

Complete step by step answer:
The reaction given to us for the dissociation of \[{N_2}{O_4}\] is as follows:
\[{N_2}{O_4} \rightleftharpoons 2N{O_{2(g)}}\]​
Initial moles
of the reactants 1 0
At equilibrium 1−α 2α
Total moles of gas at equilibrium will be = 1−α+2α = 1+α
If P is the total pressure at equilibrium, then partial pressures of the gas constituting are given by
$P_a$ = Mole fraction of gas × Total pressure
Applying this to determine the partial pressure of ${N_2}{O_4}$ , we have:
\[{P_{{N_2}{O_4}}} = \dfrac{{P(1 - \alpha )}}{{(1 + \alpha )}}\]
Similarly, for determining the partial pressure of nitrogen dioxide, we have:
\[{P_{N{O_{2(g)}}}} = \dfrac{{P(2\alpha )}}{{(1 + \alpha )}}\]​
The equilibrium constant, thus, will be determined by:
\[{K_p} = \dfrac{{{{\left[ {{P_{{N_2}{O_4}}}} \right]}^2}}}{{\left[ {{P_{N{O_{2(g)}}}}} \right]}}\]​

Substituting the values of partial pressure in the above equation, we get:
\[{K_p} = \dfrac{{{{\left( {\dfrac{{P(2\alpha )}}{{(1 + \alpha )}}} \right)}^2}}}{{\left( {\dfrac{{P(1 - \alpha )}}{{(1 + \alpha )}}} \right)}}\]
On solving further:
\[{K_p} = \dfrac{{{P^2}4{\alpha ^2}}}{{\left( {P(1 - \alpha )(1 + \alpha )} \right)}}\]


\[{K_p} = \dfrac{{P(4{\alpha ^2})}}{{\left( {1 - {\alpha ^2}} \right)}}\]
\[{K_p} - {K_p}{\alpha ^2} = 4P{\alpha ^2}\]
\[{\alpha ^2}(4P + {K_p}) = {K_p}\]​
And hence on doing the further simplification we have,
\[{\alpha ^2} = \dfrac{{{K_p}}}{{4P + {K_p}}}\]​​
\[\alpha = \sqrt {\dfrac{{{K_p}}}{{4 + {K_p}}}} \]
Hence, the degree of dissociation for the given reaction will be\[\alpha = \sqrt {\dfrac{{{K_p}}}{{4 + {K_p}}}} \]​​​

Therefore, the correct option is option D.

Note:
\[{K_p}\] is defined as the equilibrium constant and is calculated from the partial pressures of a reaction equation. It is used to express the relationship between the pressures of the product to those of the reactant. Some of the factors affecting the degree of dissociation are as follows: At normal dilution, value of is nearly 1 for strong electrolytes as there is complete dissociation of these electrolytes, while it is very less than 1 for weak electrolytes. Higher the dielectric constant of a solvent more is its ionising power and more will be the dissociation. Dilution of solution affects the amount of solvent and therefore the dissociation. It also depends on the volume. When the volume increases the equilibrium of the given reaction would shift towards the side where the gaseous molecules are more and therefore the degree of dissociation towards the side increases.