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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