The degree of dissociation of PCl5(.g.) for the equilibrium PCl5(g)leftharpoons PCl3(g)+Cl2(.g.) is approximately related to the pressure at equilibrium (P) by the relation [.α ll1].

Question

The degree of dissociation of PCl5(.g.) for the equilibrium PCl5(g)leftharpoons PCl3(g)+Cl2(.g.) is approximately related to the pressure at equilibrium (P) by the relation [.α ll1]. (A) α propto P (B) α propto (1/√P) (C) α propto (1/P2) (D) α propto (1/P4)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/374ba7f99069626e1ea6a8915abda659-.png" /> ∴ KP=(((α P/1 + α )) ((α /1 + α ) P)/((1 - α )1 + α ) P) =((α )2/1 - (α )2)P≈(α )2P(α ll 1) Neglecting ' α 2 ' KP=α 2P ∴ α =(√KP/√P)

Q. 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 $[α ≪ 1]$

$α \propto P$

$α \propto \frac{1}{P}$

$α \propto \frac{1}{P ^{2}}$

$α \propto \frac{1}{P ^{4}}$

$∴ K_{P} = \frac{( \frac{α P}{1 + α} ) ( \frac{α}{1 + α} P )}{( \frac{1 - α}{1 + α} ) P}$
$= \frac{( α ) ^{2}}{1 - ( α ) ^{2}} P \approx (α)^{2} P (α ≪ 1)$
Neglecting ' $α^{2}$ '
$K_{P} = α^{2} P$
$∴ α = \frac{K _{P}}{P}$

Q. The degree of dissociation of PCl5​(g) for the equilibrium PCl5​(g)⇌PCl3​(g)+Cl2​(g) is approximately related to the pressure at equilibrium (P) by the relation [α≪1]

α∝P

α∝P​1​

α∝P21​

α∝P41​

∴KP​=(1+α1−α​)P(1+ααP​)(1+αα​P)​ =1−(α)2(α)2​P≈(α)2P(α≪1) Neglecting ' α2 ' KP​=α2P ∴α=P​KP​​​