At 450K, `K_(p)=2.0xx10^(10)`/bar for the given reaction at equilibrium, `2SO_(2)(g)+O_(2)hArr 2SO_(3)(g)`. What is `K_(c)` at this temperature.

Explanation

For the given reaction, `Deltan=2-(2+1)=-1`.
The relation between `K_(p) and K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)`
Given: `K_(p)=2.0xx10^(10)"bar"^(-1),T=450K`
Putting the values of `K_(p)`, T and `Deltan` into the above relation, we have `K_(c)=(K_(p))/((RT)^(-1))=K_(p)RT=2xx10^(10)xx0.0821xx450`
`=7.39xx10^(11)L*mol^(-1)" "[R=0.0821L*"bar"*mol^(-1)K^-1]`.