Question

At ${727}^{0}C$ and 1.23 atm of total equilibrium pressure, $S{O}_3$ is partially dissociated into $S{O}_2$ and $O_2$ according to $S{O}_3\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+1/2O_2\left(g\right)$. The density of equilibrium mixture is 0.9 gm/litre.The degree of dissociation is

• A. 1/3
• B. 2/3
• C. 1/4
• D. 1/5

Answer 1

The correct option is B 2/3

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

We all know that,

$\alpha =\frac{D-d}{d(y-1)}$

where,

$\alpha \to$ degree of dissociation

$D\to$ Calculated vapour density

$d\to$ Observed vapour density

$y\to$ number of product per mole of reactant $=\frac{3}{2}$

$d=\frac{MolarMass}{2}$

$MolarMass=\frac{0.9\times 0.0821\times 1000}{1.23}=60$

$d=30$

$D=\frac{MolarMas{s}_{\left(S{O}_3\right)}}{2}=\frac{80}{2}=40$

$\alpha =\frac{10\times 10}{30\times 0.5}=\frac{2}{3}$

Hence, option (B) is correct.