Answer : Option B) [tex] K_{eq} = \frac{[SO_{3}]^{2} }{[SO_{3}]^{2} . [O_{3}]^{2}} [/tex]
Explanation : The equilibrium constant is often expressed in terms of the molar concentration of products divided by the product of molar concentrations of reactants.
So, [tex] K_{eq} = \frac{[products]}{[reactants]} [/tex].
Now, here in the given reaction of
[tex]2SO_{2} + O_{2} ----\ \textgreater \ 2SO_{3}[/tex]
So, we have the equation for equilibrium as [tex] K_{eq} = \frac{[SO_{3}]^{2} }{[SO_{3}]^{2} . [O_{3}]^{2}} [/tex]