Question

$S{O}_3$ decomposes at a temperature of 1000K and a total pressure of 1.642 atm. At equilibrium, density of mixture is found to be 1.28 g/l. Find the degree of dissociation of $S{O}_3$ for $S{O}_3\rightleftharpoons S{O}_2+1/2O_2.$

• A. $\alpha =0.5$
• B. $\alpha =1.5$
• C. $\alpha =0.15$
• D. $\alpha =0.8$

Answer 1

The correct option is A $\alpha =0.5$

Given:

$S{O}_3$ decomposes at a temperature of $1000K$ and a total pressure of $1,642$ atm.

At equilibrium, density of mixture $\left(\rho \right)=1.28$ g/l.

$S{O}_3\leftrightharpoons S{O}_2+1/2O_2$

Lets calculate the average molecular weight of the mixture from the density.
$\rho =\frac{PM}{RT}$
$1.28=\frac{1.642M}{0.082\times 1000}$ $⟹M=63.92$ mol/lit.

Let $\alpha$ be the degree of dissociation.
The equilibrium number of moles of $S{O}_3,S{O}_2$ and $O_2$ are $(1-\alpha ),\alpha$ and $0.5\alpha$ respectively.

The expression for the average molecular mass is $\frac{80(1-\alpha )+64\alpha +32\times 0.5(1-\alpha )}{(1-\alpha )+\alpha +0.5\alpha }=\mathrm{63.92.}$
Hence, $\alpha =\mathrm{0.5.}$