Question

Consider a reaction:
$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$
Initially only $S{O}_{2}C{l}_2\left(g\right)$ is present. The vapour density of an equilibrium mixture of this reaction is 45.
The degree of dissociation of $S{O}_{2}C{l}_2\left(g\right)$ is :

• A. 0.75
• B. 0.14
• C. 0.5
• D. 0.28

Answer 1

The correct option is C 0.5

Consider n moles of $S{O}_{2}C{l}_2$ initially.
$\alpha$ is degree of dissociation
$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)Initial:n00Equilibrium:n(1-\alpha )n\alpha n\alpha$
Total moles at equilibrium : $n(1-\alpha )+n\alpha +n\alpha =n(1+\alpha )$
Molecular weight at equilibrium = $2\times \text{Vapor Density}$ = 90
Now, according to law of conservation of mass
$\text{(Mol wt.}\times {\text{moles)}}_{initally}=\text{(Mol wt.}\times {\text{moles)}}_{equilibrium}\text{putting the values},(135\times n)=90\times n(1+\alpha )\frac{135}{90}=1+\alpha \alpha =1.50-1=0.50$

Degree of dissociation = 0.5

Alternate Solution:
we have the relation, $\alpha =\frac{(D-d)}{(n-1)d}$
where, $\alpha$ is degree of dissociation & $n$ is number of moles of product formed from one mole of reactant.
$D=\frac{\text{Molecular weight of}S{O}_{2}C{l}_2}{2}=\frac{135}{2}=67.5n=2d=\text{Vapour Density=45}$
Putting in the relation we get,
$\alpha =\frac{(67.5-45)}{(2-1)\times 45}=\frac{22.5}{45}=0.5\alpha =0.5$
Degree of dissociation = 0.5