At a certain temperature and a total pressure of `10^(5) Pa`, iodine vapour contains `40%` by volume of `I "atoms"`, Calculate `K_(p)` for the equilibrium.
`I_(2(g))hArr2I_((g))`

Explanation

`{:(,I_(2(g)),hArr,2I_((g))),("Initital moles:",a,,0),("Moles at equilibrium",(a-x),,2x):}`
`:. Total "moles" at equilibrium =a-x+2x`
`=a+x`
Given : `(2x)/(a+x)=(40)/(100)`, `:. X=(a)/(4)`
(`:'` Mole ratio= Voulme ratio, if `P` and `T` are constant)
Thus, `n_(I_(2))=a-(a)/(4)=(3a)/(4)`, `n_(1)=(a)/(2)`, `Deltan=2-1=1`
`:. K_(p)=([n_(1)]^(2))/([n_(I_(2))])xx[(P)/(sumn)]^(Delta n)`
`((a^(2)//4)/(3a//4))xx[(10^(5))/([a+(a//4)])]^(1)`
`=2.67xx10^(4)Pa`