The degree of dissociation $\mathrm{\text{'}}\mathrm{\alpha }\mathrm{\text{'}}$ of the reaction

${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{N}{\mathrm{O}}_2\left(\mathrm{g}\right)$

Can be related to ${\mathrm{K}}_{\mathrm{p}}$ as [Given: Total pressure at equilibrium = P)

• A

  $\mathrm{\alpha }=\frac{\frac{{\mathrm{K}}_{\mathrm{P}}}{\mathrm{P}}}{4+\frac{{\mathrm{K}}_{\mathrm{P}}}{\mathrm{P}}}$
• B

  $\mathrm{\alpha }=\frac{{\mathrm{K}}_{\mathrm{P}}}{4+{\mathrm{K}}_{\mathrm{P}}}$
• Correct answer
  C

  $\mathrm{\alpha }={\left[\frac{{\mathrm{K}}_{\mathrm{P}}/\mathrm{P}}{4+{\mathrm{K}}_{\mathrm{P}}/\mathrm{P}}\right]}^{1/2}$
• D

  $\mathrm{\alpha }={\left[\frac{{\mathrm{K}}_{\mathrm{P}}}{4+{\mathrm{K}}_{\mathrm{P}}}\right]}^{1/2}$

Solution

${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{N}{\mathrm{O}}_2\left(\mathrm{g}\right)$

${\mathrm{K}}_{\mathrm{P}}=\frac{{\left(2\mathrm{\alpha }\right)}^2}{\left(1–\mathrm{\alpha }\right)}\times {\left(\frac{\mathrm{P}}{1+\mathrm{\alpha }}\right)}^1$
${\mathrm{K}}_{\mathrm{P}}=\frac{4{\mathrm{\alpha }}^2}{1–{\mathrm{\alpha }}^2}\mathrm{P}$
$\mathrm{\alpha }={\left[\frac{{\mathrm{K}}_{\mathrm{P}}/\mathrm{P}}{4+{\mathrm{K}}_{\mathrm{P}}/\mathrm{P}}\right]}^{1/2}$

Hence, the correct answer is option: 'C'.