Consider the reaction  ${\mathrm{X}}_2\mathrm{Y}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{X}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Y}}_2\left(\mathrm{g}\right)$

The equation representing the correct relationship between the degree of dissociation (x) of X₂Y(g) with its equilibrium constant Kp is________.
Assume x to be very, very small.

1 mole

$\begin{array}{l}{\mathrm{X}}_2{\mathrm{Y}}_{\left(\mathrm{g}\right)} \rightleftharpoons {\mathrm{X}}_{2\left(\mathrm{g}\right)}+ \frac{1}{2}{\mathrm{Y}}_{2\left(\mathrm{g}\right)}\\ \text{1-x mole }\mathrm{x}\text{ mole }\frac{\mathrm{x}}{2}\text{ mole }\end{array}$

$\begin{array}{l}\therefore {\mathrm{P}}_{{\mathrm{X}}_2\mathrm{Y}}=\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\times \mathrm{P}\\ {\mathrm{P}}_{{\mathrm{X}}_2}=\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\times \mathrm{P}\\ {\mathrm{P}}_{{\mathrm{Y}}_2}=\frac{\mathrm{x}/2}{1+\frac{\mathrm{x}}{2}}\times \mathrm{P}\end{array}$

$\begin{array}{l}\therefore {\mathrm{K}}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\mathrm{P}{\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)}\mathrm{P}\right)}^{\frac{1}{2}}/\left(\frac{1-\mathrm{x}}{1+\frac{\mathrm{x}}{2}}\right)\times \mathrm{P}\right.\\ \therefore {\mathrm{K}}_{\mathrm{p}}=\left(\frac{\mathrm{x}}{1-\mathrm{x}}\right){\left(\frac{\mathrm{x}}{2\left(1+\frac{\mathrm{x}}{2}\right)}\right)}^{\frac{1}{2}}\times {\mathrm{p}}^{\frac{1}{2}}\end{array}$

$\begin{array}{r}\because \mathrm{x}\text{ to be very very small }\\ \begin{array}{rl} & \therefore {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{x}}^{3/2}}{(2{)}^{\frac{1}{2}}}\times {\mathrm{P}}^{\frac{1}{2}}\\ & \therefore {\mathrm{x}}^{\frac{3}{2}}=\frac{{\mathrm{K}}_{\mathrm{p}}\times 2^{\frac{1}{2}}}{{\mathrm{P}}^{\frac{1}{2}}}\\ & \therefore {\mathrm{x}}^3=\frac{{\mathrm{K}}_{\mathrm{p}}^2\times 2}{\mathrm{P}}\\ & \mathrm{x}={\left(\frac{{\mathrm{K}}_{\mathrm{p}}^2\times 2}{\mathrm{P}}\right)}^{\frac{1}{3}}\end{array}\end{array}$