At a certain temperature & total pressure of `10^(5)Pa`, iodine vapour contians 40% by volume 1 atoms, `I_(2)(g)hArr 2I(g)`. Calculate `K_(p)` for the equilibrium.

Explanation

`I_(2)(g)hArr 2I(g)`, In the mixture, volume fraction of `I(g)=0.4 and` that of `I_(2)(g)=1-0.4=0.6`
We know, volume fraction=mole fraction
Therefore, in the mixture the mole fraction of I(g) and `I_(2)(g)` are 0.4 and 0.6 respectively and their partial pressures are `p_(1)=x_(1)xxP=0.4xx10^(5)Pa=4xx10^(4)Pa and p_(I_(2))xxP=0.6xx10^(5)=6xx10^(4)Pa`
`therefore K_(p)=(p_(1)^(2))/(p_(I_(2)))=((4xx10^(4))^(2)Pa)/(6xx10^(4))=2.66xx10^(4)Pa`.