Question

At ${627^ \circ }C$ and $1atm$, $S{O_3}$ is partially dissociated into $S{O_2}$ and ${O_2}$.$S{O_3}_{\left( g \right)} \leftrightarrow S{O_2}_{\left( g \right)} + \dfrac{1}{2}{O_2}_{\left( g \right)}$ . If the density of the equilibrium mixture is, then what will be the degree of dissociation?
A.$0.74$
B.$0.34$
C.$0.68$
D.$0.64$

Answer 1

Hint:Degree of dissociation is the fraction of molecules dissociating in a given time. It is denoted by $\alpha $ .
Formula used:The relation between vapor density and degree of dissociation is given by the formula: $\therefore \alpha = \dfrac{{D - d}}{{\left( {n - 1} \right)d}}$where, $\alpha = $ degree of dissociation, $D = $ initial vapor density, $d = $ vapor density at equilibrium, $n = $ number of moles.

Complete step by step answer:
Degree of dissociation is defined as the fraction of molecules which dissociate in a given time. It is denoted by the symbol $\left( \alpha \right)$ .
It is given by the formula: ${K_a} = \dfrac{{{\alpha ^2}C}}{{1 - \alpha }}$
Where, ${K_a} = $ acid dissociation constant
$C = $ Concentration
$\alpha = $ degree of dissociation of acid.
A.Given data:
Temperature: ${627^ \circ }C$
$T = 627 + 273$
$T = 900K$
Pressure: $1atm$
Reaction: $S{O_3}_{\left( g \right)} \leftrightarrow S{O_2}_{\left( g \right)} + \dfrac{1}{2}{O_2}_{\left( g \right)}$
Density of the equilibrium mixture = $0.925g/{L^{ - 1}}$
B.Now we will see the molecular mass of the mixture at equilibrium which will be denoted by ${M_{mix}}$.
By applying the relation,
${M_{mix}} = \dfrac{{dRT}}{P}$
Where,
${M_{mix}} = $ molecular mass of the mixture
$d = $ density
$R = $ gas constant
$T = $ Temperature
$P = $ Pressure.
Using this formula we are going to find the molecular mass of the mixture
${M_{mix}} = \dfrac{{dRT}}{P}$
Substituting the values we get,
${M_{mix}} = \dfrac{{0.925 \times 0.0821 \times 900}}{1}$
${M_{mix}} = 68.348$ .
C) After finding the molecular mass of the mixture, we will find the vapour density of the mixture, d
Vapour density of the mixture$d = \dfrac{{68.34}}{2}$
$d = 34.17$
Molecular mass of $S{O_3} = 80$
So, the vapour density of $S{O_3},D = \dfrac{{80}}{2}$
$\therefore D = 40$ .
D) Relation between degree of dissociation and vapor densities:
Let us consider the reaction: $A \rightleftharpoons nB$
Initially $c = 0$, at equilibrium : $c\left( {1 - \alpha } \right) = nc\alpha $
Therefore at equilibrium:$c\left[ {1 + \alpha \left( {n - 1} \right)} \right]$
Now let the initial vapor density be $D$ and vapor density at equilibrium be $d$ .
$\dfrac{{initial}}{{equilibrium}} = \dfrac{{total moles}}{{initial moles}}$
$\dfrac{D}{d} = \dfrac{{c\left[ {1 + \alpha \left( {n - 1} \right)} \right]}}{c}$
$\therefore \alpha = \dfrac{{D - d}}{{\left( {n - 1} \right)d}}$
Now finally after getting the values for vapour densities of mixture and $S{O_3}$ we will now calculate the degree of dissociation
Let the degree of dissociation be $'x'$ .
$x = \dfrac{{D - d}}{{\left( {n - 1} \right)d}}$
Where ,
$D = $ vapor density of $S{O_3}$
$d = $ vapor density of the mixture
Substituting the values we get,
$x = \dfrac{{40 - 34.17}}{{(\dfrac{3}{2} - 1) \times 34.17}}$
$x = \dfrac{{5.82 \times 2}}{{34.17}}$
$x = 0.34$
So, the correct answer will be option B) $0.34$.

Note:
Dissociation of ions occurs only when the solid ionic dissolves completely. The nonionic compound does not dissociate in water.so be careful while solving sums related to this concept.