Question

For the reaction ${N_2}{O_4} \rightleftarrows 2N{O_{2(g)}}$ the degree of dissociation is $0.2$ at equilibrium and $1atm$ pressure then equilibrium constant ${K_p}$ will be?

Answer 1

Hint: Degree of dissociation is basically the amount of solute which is dissociated into its ion form or radical form per mole. The degree of dissociation is denoted by the symbol, $\alpha $. This depends upon several factors such as the nature of solvent and solute, temperature, and concentration.

Complete step by step answer:
In the question we are given a reaction of decomposition of ${N_2}{O_4}$, that is, ${N_2}{O_4} \rightleftarrows 2N{O_{2(g)}}$.
So as we can see ${N_2}{O_4}$ is getting dissociated into two moles of $N{O_2}$, so we can assume:
${N_2}{O_4} \rightleftharpoons 2N{O_2}$
    $1$ $0$ $ - - - $at $t = 0$
$1 - \alpha $ $2\alpha $ $ - - - $ at $t = t$
So, we have assumed at $t = 0$, the concentration of ${N_2}{O_4}$ is $1$ and the concentration of $N{O_2}$ is $0$ and at $t = t$, the concentrations of ${N_2}{O_4}$ and $N{O_2}$ are $1 - \alpha $ and $2\alpha $.
Now we will calculate the total moles:
Total moles $ = 1 - \alpha + 2\alpha = 1 + \alpha $
Since, we are given pressure, then we need to find out the answer in terms of partial pressure only. The formula which we will use is:
Partial Pressure $ = \dfrac{{1 - \alpha }}{{1 + \alpha }} \times P\dfrac{{2\alpha }}{{1 + \alpha }} \times P$, where P is the pressure and $\alpha $ is the degree of dissociation.
As we know the values of $\alpha = 0.2$ and $P = 1$, so we will put these values in the above formula to find out the partial pressure of ${N_2}{O_4}$ and $N{O_2}$.
${p_{{N_2}{O_4}}}$$ = \dfrac{{1 - 0.2}}{{1 + 0.2}} \times 1 = \dfrac{{0.8}}{{1.2}} = \dfrac{2}{3}$
${p_{N{O_2}}} = \dfrac{{2 \times 0.2}}{{1 + 0.2}} \times 1 = \dfrac{{0.4}}{{1.2}} = \dfrac{1}{3}$
Now, we have the values for the partial pressure of ${N_2}{O_4}$ as well as $N{O_2}$, so we can easily find out the equilibrium constant, ${K_p}$ by dividing the square of both the partial pressures:
$\dfrac{1}{6}$${K_p} = \dfrac{{{{[{p_{N{O_2}}}]}^2}}}{{{{[{p_{{N_2}{O_4}}}]}^2}}} = \dfrac{{1 \times 1 \times 3}}{{3 \times 3 \times 2}} = \dfrac{1}{6}$
Therefore, the equilibrium constant, ${K_p}$ is $\dfrac{1}{6}$.

Note:
${K_p}$ is known as the equilibrium constant which can be calculated by dividing the partial pressures of the reactants and the products, the way we have done in the above solution. The equilibrium constant can also be defined as the relation between the partial pressure of the reactants and the products formed.