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. CaCO_(3)(s)hArr+CaO(s)+CO_(2)(s) At equilibrium in the above case, 'a' moles of CaCO_(3), 'b' moles of CaO and 'c' moles of CO_(2) are found then identify the wrong statement: