Activation energy, Standard free energy and Degree of dissociation and Vapour density
Category : NEET
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
You need to login to perform this action.
You will be redirected in
3 sec