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Question: \( S{{O}_{3}} \) decomposes at a temperature of \( 1000\text{ }K \) and a total pressure of \( 1.642...

SO3S{{O}_{3}} decomposes at a temperature of 1000 K1000\text{ }K and a total pressure of 1.642 atm.1.642\text{ }atm. At equilibrium, density of mixture is found to be 1.28g/l.1.28g/l. Find the degree of dissociation of SO3S{{O}_{3}} for SO3SO2+12O2.S{{O}_{3}}S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}.

Explanation

Solution

Hint : We know that the 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. We have to use here the relation M=ρRTP.M=\rho RTP.

Complete Step By Step Answer:
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=ρRTP.M=\rho RTP.
Here, M is the molar mass, is the density of the gas, P is the pressure, T is the temperature and R is the universal gas constant. In the question, we have SO3S{{O}_{3}} decomposes at a temperature of 1000 K 1000~K~ and a total pressure of 1,642 atm.1,642~atm. At equilibrium, density of mixture (ρ)=1.28 g/l.\left( \rho \right)=1.28~g/l.
SO3SO2+12O2.S{{O}_{3}}S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}.
Lets calculate the average molecular weight of the mixture from the density. M=ρRTP.M=\rho RTP.
∴ρ=MPRT\therefore \rho =\dfrac{MP}{RT}
Substituting all the values; ∴ρ=M×1.6420.082×1000⇒M=63.92mol/lit.\therefore \rho =\dfrac{M\times 1.642}{0.082\times 1000}\Rightarrow M=63.92mol/lit.
Let α\alpha be the degree of dissociation.
The equilibrium number of moles of SO3, SO2S{{O}_{3}},~S{{O}_{2}} and O2{{O}_{2}} ​ are (1−α),α \left( 1-\alpha \right),\alpha ~ and 0.5 α0.5~\alpha respectively.
The expression for the average molecular mass is [80(1−α)]+[64α]+[32×0.5(1−α)][(1−α)+α+0.5α]=63.92\dfrac{\left[ 80(1-\alpha ) \right]+\left[ 64\alpha \right]+\left[ 32\times 0.5(1-\alpha ) \right]}{\left[ (1-\alpha )+\alpha +0.5\alpha \right]}=63.92
On further simplifying we get degree of dissociation (α)\left( \alpha \right) is 0.5.0.5.

Note :
Remember that 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 Ksp.Ksp.