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. `K_(p)` for the reaction `NH_(4)I(s)hArrNH_(3)(g)+HI(g)`is `1//4 at 300K`.If above equilibrium is established by taking 4 moles of `NH_(4)I(s)` in 100 litre contanier, then moles of `NH_(4)I(s)` left in the container at equilibrium is `["Taken R=1/12Lt.atm mol"^(-1)K^(-1)]`.A. 1B. 2C. 3D. 4