If more than one phase is present in the reversible reaction then it is said to be heterogenous system. Example: `CaO(s)+CO_(2)(g)` Expression of equilibrium constant for the above reaction can be taken as : `K=([CaO(s)][CO_(2)(g)])/([CaO(s)])`." ".....(i) Now concentration of `CaO(s)=[CaO(s)]` `=("moles of CaO")/("volume of CaO")` as density of `CaO[rho_(CaO(s))]` and molar mass of `CaO[M_(CaO(s))]`are a fixed quantity therefore concentration of pure solid and liquid term is uncharge with respect to time. Hence, equilibrium constant for the equation (i) can be written as : `K_(C)=[CO_(2)(g))]` `K_(P)=P_(CO_2)` As `K_(p) and K_(c)` is not containing solid terms therefore, addition or removel of pure solid and pure liquid has no effect on the equilibrium process. `200g of CaCO_(3)(g)` taken in 4Ltr container at a certain temperature. `K_(c)` for the dissociation of `CaCO_(3)` at this temperature is found to be `1//4` mole `Ltr^(-1)`then the concentration of CaO in mole/litre is : [Given :`rho