The degree of dissociation of PCl5(.α .) obeying the equilibrium, PCl5(.g.)leftharpoons PCl3(.g.)+Cl2(.g.) , is approximately related to the pressure at equilibrium by

Question

The degree of dissociation of PCl5(.α .) obeying the equilibrium, PCl5(.g.)leftharpoons PCl3(.g.)+Cl2(.g.) , is approximately related to the pressure at equilibrium by (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="Solution" src="https://cdn.tardigrade.in/q/nta/c-l7csz83ovbat.png" /> So, mathrmK mathrmp=(α2/(1-α))[( mathrmP/1+α)]=(α2 mathrmP/1-α2) or α =√(KP/P) (if 1-α 2=1 .)

Q. The degree of dissociation of $PC l_{5} (α)$ obeying the equilibrium,

$PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$ , is approximately related to the pressure at equilibrium by

$α \propto P$

$α \propto \frac{1}{P}$

$α \propto \frac{1}{P ^{2}}$

$α \propto \frac{1}{P ^{4}}$

So, $K_{p} = \frac{α ^{2}}{( 1 - α )} [\frac{P}{1 + α}] = \frac{α ^{2} P}{1 - α ^{2}}$

or $α = \frac{K _{P}}{P}$ (if $1 - α^{2} = 1$ .)

Q. The degree of dissociation of PCl5​(α) obeying the equilibrium, PCl5​(g)⇌PCl3​(g)+Cl2​(g) , is approximately related to the pressure at equilibrium by

α∝P

α∝P​1​

α∝P21​

α∝P41​

So, Kp​=(1−α)α2​[1+αP​]=1−α2α2P​ or α=PKP​​​ (if 1−α2=1 .)