The degree of dissociation of PC l _ 5 ( g ) for the equilibrium PC l _ 5 (g) ⇌ PC l _ 3 (g) + C l _ 2 ( g ) is approximately related to the pressure at equilibrium (P) by the relation [ α ≪ 1 ]

Correct answer: B. α ∝ 1 P cases PCl _ 5 & ⇌ & PCl _ 3 & + & Cl _ 2 & (Total pressure = P) 1 & & 0 & & 0 & 1 - α & & α & & α & cases ∴ K _ P = ((( )/(1 + α))((α)/(1 + α)P))/(((1 - α)/(1 + α))P) = α ^ 2 1 - α ^ 2 P α ^ 2 P(α ≪ 1) Neglecting ' α ^ 2 ' K _ P = α ^ 2 P ∴ α = K _ P P

NTA Abhyas 2020

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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The degree of dissociation of PC l _ 5 ( g ) for the equilibrium PC...

The degree of dissociation of $PC{l}_{5} ( g )$ for the equilibrium $PC{l}_{5}\left(g\right) ⇌ PC{l}_{3}\left(g\right) + C{l}_{2} ( g )$ is approximately related to the pressure at equilibrium (P) by the relation $[ \alpha ≪ 1 ]$

Options

A $\alpha ∝ P$
B $\alpha ∝ \dfrac{1}{\sqrt{P}}$
C $\alpha ∝ \dfrac{1}{P^{2}}$
D $\alpha ∝ \dfrac{1}{P^{4}}$

Correct answer

B. $\alpha ∝ \dfrac{1}{\sqrt{P}}$

Step-by-step solution

\begin{cases}{\text{PCl}}_{5} & ⇌ & {\text{PCl}}_{3} & + & {\text{Cl}}_{2} & \left(\text{Total pressure} = \text{P}\right) \\ 1 & & 0 & & 0 & \\ 1 - \alpha & & \alpha & & \alpha & \end{cases}

∴ {K}_{P} = \dfrac{\left(\dfrac{\alphaP}{1 + \alpha}\right)\left(\dfrac{\alpha}{1 + \alpha}P\right)}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P}

= \dfrac{{\alpha}^{2}}{1 - {\alpha}^{2}}P \approx {\alpha}^{2}P\left(\alpha ≪ 1\right)

Neglecting '

{\alpha}^{2}

{K}_{P} = {\alpha}^{2}P

∴ \alpha = \dfrac{\sqrt{{K}_{P}}}{\sqrt{P}}

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