NEET Chemistry Equilibrium / साम्यावस्था Activation energy, Standard free energy and Degree of dissociation and Vapour density

Activation energy, Standard free energy and Degree of dissociation and Vapour density

Relation between vapour density and degree of dissociation.

In the following reversible chemical equation.

                        \[A\]   ?    \[yB\]

Initial mol          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,  \[D=\frac{molecular\ weight}{volume}=\frac{m}{V}\]

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) \[2N{{O}_{2}}\rightleftharpoons {{N}_{2}}{{O}_{4}},\ y=\frac{1}{2}\]

            \[\therefore \] \[x=\frac{D-d}{d}\] (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