The degree of dissociation 'α' of the reaction N _ 2 O _ 4 (g) ⇌ 2N O _ 2 (g) Can be related to K _ p as [Given: Total pressure at equilibrium = P)

Correct answer: C. α = [ K _ P / P 4 + K _ P / P ] ^ 1 / 2 N _ 2 O _ 4 (g) ⇌ 2N O _ 2 (g) K _ P = (2α) ^ 2 (1 – α) × ((P)/(1 + α)) ^ 1 K _ P = 4 α ^ 2 1 – α ^ 2 P α = [ K _ P / P 4 + K _ P / P ] ^ 1 / 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The degree of dissociation 'α' of the reaction N _ 2 O _ 4 (g) ⇌ 2N...

The degree of dissociation

'\alpha'

of the reaction

{N}_{2}{O}_{4} \left(g\right) ⇌ 2N{O}_{2} \left(g\right)

Can be related to

{K}_{p} as [Given: Total pressure at equilibrium = P)

Options

A $\alpha = \dfrac{\dfrac{{K}_{P}}{P}}{4 + \dfrac{{K}_{P}}{P}}$
B $\alpha = \dfrac{{K}_{P}}{4 + {K}_{P}}$
C $\alpha = {\left[\dfrac{{K}_{P} / P}{4 + {K}_{P} / P}\right]}^{1 / 2}$
D $\alpha = {\left[\dfrac{{K}_{P}}{4 + {K}_{P}}\right]}^{1 / 2}$

Correct answer

C. $\alpha = {\left[\dfrac{{K}_{P} / P}{4 + {K}_{P} / P}\right]}^{1 / 2}$

Step-by-step solution

{N}_{2}{O}_{4}\left(g\right) ⇌ 2N{O}_{2}\left(g\right)

{K}_{P} = \dfrac{{\left(2\alpha\right)}^{2}}{\left(1 – \alpha\right)} \times {\left(\dfrac{P}{1 + \alpha}\right)}^{1}

{K}_{P} = \dfrac{4{\alpha}^{2}}{1 – {\alpha}^{2}}P

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

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