Question

At 672${}^{\circ }C$ and one atmosphere, $S{O}_3$ is partially dissociated into $S{O}_2$ and $O_2.$ $S{O}_{3\left(g\right)}\leftrightharpoons S{O}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$ . If the density of the equilibrium mixture is 0.925 g/litre, then what is the degree of dissociation?

• A. $0.74$
• B. $0.34$
• C. $0.68$
• D. $0.64$

Answer 1

Answer: (B)

$0.34$

$M=\frac{\rho RT}{P}$

$M=$ Molar mass of an ideal gas.

$P=$ Density at pressure $P$ and temperature $T$.

$M=\frac{0.925\times 0.021\times (627+273)}{1}$

$M=68.36g/mol$

${2S{O}_3}_{\left(g\right)}\rightleftharpoons {2S{O}_2}_{\left(g\right)}+{O_2}_{\left(g\right)}$

$1-x$ $x$ $\frac{x}{2}$

$1-x+x+\frac{x}{2}=1+\frac{x}{2}$

$\Rightarrow$ Mole fraction of $S{O}_3=\frac{1-x}{1+\frac{x}{2}}=\frac{1-x}{1+\frac{x}{2}}$

$X_{S{O}_2}=\frac{x}{1+\frac{x}{2}}$; $X_{O_2}=\frac{\frac{x}{2}}{1+\frac{x}{2}}$

$\Rightarrow 68.36=\left(\frac{1-x}{1+\frac{x}{2}}\right)80.066+\frac{x}{(1+\frac{x}{2})64.066}+\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)32.0$

From that we will get $x=0.35$.

$\therefore$ Degree of dissociation $\alpha =0.35$ .