Question

At ${{627}^{\circ }}C$ and one atmosphere pressure $S{{O}_{3}}$ is partially dissociated into $S{{O}_{2}}$ and ${{O}_{2}}$ by $S{{O}_{3}}\left( g \right)\rightleftharpoons \dfrac{1}{2}{{O}_{2}}\left( g \right)$ . The density of the equilibrium mixture is 0.925g/litre. What is the degree of dissociation?

Answer 1

Hint: To solve this question, firstly you have to calculate the number of moles of sulphur trioxide dissociated to produce sulphur dioxide and oxygen gas. For finding it out, you will need to calculate the average molar mass of the mixture. You can use the relation- $M=\rho RTP$.

Complete step by step solution:
To answer this first we should know that degree of dissociation is the extent of the molecules breaking into ions in a solution.
Now let us discuss the question given to us.
Here firstly we will find the average molar mass of the mixture by using the given density, pressure and temperature using the relation-
$M=\rho RTP$
Here, M is the molar mass, $\rho $ is the density of the gas, P is the pressure, T is the temperature and R is the universal gas constant.
In the question, temperature is given to us as ${{627}^{\circ }}C$ i.e.$ (627 + 273)K$ $= 900K$.
The pressure is given as $1atm$.
Density is given as $0.925g/litre$
So, putting these values in the above relation we can write that-
\[M=0.925\text{ }g/litre\times 0.082\text{ }Latm/molK\times 1\text{ }atm\times 900\text{ K}\]
Calculating it we will get, $M =$ $68.265 g/mol$
This is the average molar mass of the mixture of gases. We can find the contribution of each gas by dividing its mole fraction by the average molar mass. We can write the balanced equation of the reaction as-
\[2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)\]
From the above reaction, we can understand that 2 moles of sulphur trioxide dissociate to give 2 moles of sulphur dioxide and one mole of oxygen gas. Let us assume that the degree of dissociation is ‘x’ and we have one mole of sulphur dioxide. So, we can write that-

	$S{{O}_{3}}$	$S{{O}_{2}}$	${{O}_{2}}$
Moles before dissociation	1	0	0
Moles after dissociation	(1-x)	x	$\dfrac{x}{2}$

So, at equilibrium the number of moles of gas present in total will be = $\left( 1-x+x+\dfrac{x}{2} \right)=\left( 1+\dfrac{x}{2} \right)$
Now, we can write that mole fraction of sulphur trioxide = $\dfrac{1-x}{1+\dfrac{x}{2}}$ and that of sulphur dioxide and oxygen will be $\dfrac{x}{1+\dfrac{x}{2}}$ and $\dfrac{\dfrac{x}{2}}{1+\dfrac{x}{2}}$ respectively.
We know that molar mass of sulphur dioxide, sulphur trioxide and oxygen is $64.066 g/mol$, $80.066g/mol$ and $32 g/mol$ respectively.
But, we have calculated above that the average molar mass is $68.265 g/mol$.
Therefore, we can write that-
$\left[ \dfrac{1-x}{1+\dfrac{x}{2}} \right]\times 80.066\text{ + }\left[ \dfrac{x}{1+\dfrac{x}{2}} \right]\times 64.066\text{ + }\left[ \dfrac{\dfrac{x}{2}}{1+\dfrac{x}{2}} \right]\times 32\text{ = }68.265$
Now, calculating for ‘x’ from the above equation we will get that $x = 0.35$.
And we have assumed that x is the degree of dissociation i.e. for $1$ mole of sulphur trioxide, $0.35$ moles of it will dissociate into sulphur dioxide and oxygen gas.

Therefore, the correct answer is $0.35$.

Note: Dissociation is the separation of ions that occurs when a solid ionic compound dissolves. Non-ionic compounds do not dissociate in water. Be very careful of this distinction so as to ensure you make no silly mistakes when trying to solve questions related to this concept. Measure of degree of dissociation is ${{K}_{sp}}$. It is written in terms of concentration of the ions- \[Ksp=\left[ cation \right]\times \left[ anion \right]\]