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

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.