Question
Question: One mole of \(S{{O}_{3}}\) was placed in a two-litre vessel at a certain temperature. The following ...
One mole of SO3 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)
The equilibrium mixture reacted with 0.2mole KMnO4 in acidic medium. Hence, KC is:
A. 0.50
B. 0.25
C. 0.125
D. None of these
Solution
First of all, know about what is the charge change for the KMnO4 and SO2. 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 SO3 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)
The equilibrium mixture was further reacted with 0.2mole KMnO4 in acidic medium. And we have to find out the equilibrium constant.
We should know that, when we react the equilibrium mixture with KMnO4, the SO2 present in the mixture gets oxidized and two moles of KMnO4 will oxidize five moles of SO2. The equivalent of SO2 and KMnO4 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 SO3, SO2 and O2 are 1, 0 and 0 respectively.
While, the equilibrium number of moles of SO3, SO2 and O2 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 KMnO4 and SO2, we get:
2X×2=0.2×5
Then, X=0.25
So, the equilibrium number of moles of SO3, SO2 and O2 will be 1−2X=1−2×0.25=0.5, 2X=2×0.25=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 SO3, SO2 and O2 will be 20.5=0.25M, 20.5=0.25M and 20.25=0.125M respectively.
By using the formula for equilibrium constant (KC) 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 will be:
KC=[SO3]2[SO2]2[O2]
Then, KC=[0.25]2[0.25]2×0.125=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.