# JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Relation Between Vapour Density And Degree Of Dissociation

Relation Between Vapour Density And Degree Of Dissociation

Category : JEE Main & Advanced

In the following reversible chemical equation.

$A$  $\rightleftharpoons$   $yB$

Initial mole              1              0

At equilibrium  (1–x)                   yx        x = degree of dissociation

Number of moles of $A$ and $B$ at equilibrium $=1-x+yx=1+x(y-1)$

If initial volume of 1 mole of A is V, then volume of equilibrium mixture of $A$ and $B$ is,$=[1+x(y-1)]V$

Molar density before dissociation,

$D=\frac{\text{molecular}\ \text{weight}}{\text{volume}}=\frac{m}{V}$

Molar density after dissociation, $d=\frac{m}{[1+x(y-1)]V}$;$\frac{D}{d}=[1+x(y-1)]$ ; $x=\frac{D-d}{d(y-1)}$

$y$ is the number of moles of products from one mole of reactant. $\frac{D}{d}$ is also called Van’t Hoff factor.

In terms of molecular mass,$x=\frac{M-m}{(y-1)\,m}$

Where $M=$ Initial molecular mass,

$m=$ molecular mass at equilibrium

Thus for the equilibria

(I) $PC{{l}_{5(g)}}$ $\rightleftharpoons$  $PC{{l}_{3(g)}}+C{{l}_{2(g)}},y=2$

(II) ${{N}_{2}}{{O}_{4(g)}}$ $\rightleftharpoons$  $2N{{O}_{2(g)}},\ y=2$

(III) $2N{{O}_{2}}$ ? ${{N}_{2}}{{O}_{4}},\ y=\frac{1}{2}$

$\therefore$ $x=\frac{D-d}{d}$ (for I and II)  and $x=\frac{2(d-D)}{d}$ (for III)

Also $D\times 2=$ Molecular weight (theoretical value)

$d\times 2=$ Molecular weight (abnormal value) of the mixture.

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