Question

One mole of $S{{O}_{3}}$ was placed in a two-litre vessel at a certain temperature. The following equilibrium was established in the vessel:
$2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)$
The equilibrium mixture reacted with $0.2$mole $KMn{{O}_{4}}$ in acidic medium. Hence, ${{K}_{C}}$ is:
A. $0.50$
B. $0.25$
C. $0.125$
D. None of these

Answer 1

Hint: First of all, know about what is the charge change for the $KMn{{O}_{4}}$ and $S{{O}_{2}}$. And find out the initial number of moles and the number of moles in equilibrium. Then, find out the equilibrium concentrations of the products and the reactants. Further use the formula for the equilibrium constant.

Complete step by step solution:
Given that,
One mole of $S{{O}_{3}}$ was placed in a two-litre vessel at a certain temperature. The following equilibrium was established in the vessel:
$2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)$
The equilibrium mixture was further reacted with $0.2$mole $KMn{{O}_{4}}$ in acidic medium. And we have to find out the equilibrium constant.
We should know that, when we react the equilibrium mixture with $KMn{{O}_{4}}$, the $S{{O}_{2}}$ present in the mixture gets oxidized and two moles of $KMn{{O}_{4}}$ will oxidize five moles of $S{{O}_{2}}$. The equivalent of $S{{O}_{2}}$ and $KMn{{O}_{4}}$ should be equal and we know that equivalent equals to the product of number of moles and the valency factor.
Here, the initial number of moles of $S{{O}_{3}}$, $S{{O}_{2}}$ and ${{O}_{2}}$ are $1$, $0$ and $0$ respectively.
While, the equilibrium number of moles of $S{{O}_{3}}$, $S{{O}_{2}}$ and ${{O}_{2}}$ are $1-2X$, $2X$ and $X$, where X is the dissociation constant which is multiplied by the volume of the vessel.
So, by equating the equivalents of $KMn{{O}_{4}}$ and $S{{O}_{2}}$, we get:
$2X\times 2=0.2\times 5$
Then, $X=0.25$
So, the equilibrium number of moles of $S{{O}_{3}}$, $S{{O}_{2}}$ and ${{O}_{2}}$ will be $1-2X=1-2\times 0.25=0.5$, $2X=2\times 0.25=0.5$ and $X=0.25$ respectively.
And we know, the equilibrium concentration of a component equals the ratio of the equilibrium number of moles to that of volume of the solution.
So, the equilibrium concentration of $S{{O}_{3}}$, $S{{O}_{2}}$ and ${{O}_{2}}$ will be $\dfrac{0.5}{2}=0.25M$, $\dfrac{0.5}{2}=0.25M$ and $\dfrac{0.25}{2}=0.125M$ respectively.
By using the formula for equilibrium constant (${{K}_{C}}$) i.e. the ratio of the product of the equilibrium concentrations of the products to that of the product of the equilibrium concentrations of the reactants.
So, here the value of ${{K}_{C}}$ will be:
${{K}_{C}}=\dfrac{{{\left[ S{{O}_{2}} \right]}^{2}}\left[ {{O}_{2}} \right]}{{{\left[ S{{O}_{3}} \right]}^{2}}}$
Then, ${{K}_{C}}=\dfrac{{{\left[ 0.25 \right]}^{2}}\times 0.125}{{{\left[ 0.25 \right]}^{2}}}=0.125$

Hence, the correct option is C.

Note: It is important to note that the equilibrium constant of a chemical reaction generally provides us the relationship between the products and the reactants when the reaction reaches equilibrium and thus it says about the extent of a reaction.