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Question: One mole of \(S{{O}_{3}}\) was placed in a two-litre vessel at a certain temperature. The following ...

One mole of SO3S{{O}_{3}} was placed in a two-litre vessel at a certain temperature. The following equilibrium was established in the vessel:
2SO3(g)2SO2(g)+O2(g)2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)
The equilibrium mixture reacted with 0.20.2mole KMnO4KMn{{O}_{4}} in acidic medium. Hence, KC{{K}_{C}} is:
A. 0.500.50
B. 0.250.25
C. 0.1250.125
D. None of these

Explanation

Solution

First of all, know about what is the charge change for the KMnO4KMn{{O}_{4}} and SO2S{{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 SO3S{{O}_{3}} was placed in a two-litre vessel at a certain temperature. The following equilibrium was established in the vessel:
2SO3(g)2SO2(g)+O2(g)2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)
The equilibrium mixture was further reacted with 0.20.2mole KMnO4KMn{{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 KMnO4KMn{{O}_{4}}, the SO2S{{O}_{2}} present in the mixture gets oxidized and two moles of KMnO4KMn{{O}_{4}} will oxidize five moles of SO2S{{O}_{2}}. The equivalent of SO2S{{O}_{2}} and KMnO4KMn{{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 SO3S{{O}_{3}}, SO2S{{O}_{2}} and O2{{O}_{2}} are 11, 00 and 00 respectively.
While, the equilibrium number of moles of SO3S{{O}_{3}}, SO2S{{O}_{2}} and O2{{O}_{2}} are 12X1-2X, 2X2X and XX, where X is the dissociation constant which is multiplied by the volume of the vessel.
So, by equating the equivalents of KMnO4KMn{{O}_{4}} and SO2S{{O}_{2}}, we get:
2X×2=0.2×52X\times 2=0.2\times 5
Then, X=0.25X=0.25
So, the equilibrium number of moles of SO3S{{O}_{3}}, SO2S{{O}_{2}} and O2{{O}_{2}} will be 12X=12×0.25=0.51-2X=1-2\times 0.25=0.5, 2X=2×0.25=0.52X=2\times 0.25=0.5 and X=0.25X=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 SO3S{{O}_{3}}, SO2S{{O}_{2}} and O2{{O}_{2}} will be 0.52=0.25M\dfrac{0.5}{2}=0.25M, 0.52=0.25M\dfrac{0.5}{2}=0.25M and 0.252=0.125M\dfrac{0.25}{2}=0.125M respectively.
By using the formula for equilibrium constant (KC{{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 KC{{K}_{C}} will be:
KC=[SO2]2[O2][SO3]2{{K}_{C}}=\dfrac{{{\left[ S{{O}_{2}} \right]}^{2}}\left[ {{O}_{2}} \right]}{{{\left[ S{{O}_{3}} \right]}^{2}}}
Then, KC=[0.25]2×0.125[0.25]2=0.125{{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.