Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression ?

Explain why pure liquids and solids can be ignored while writing the e

1.	Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression ?
Answer» Solution :The molar concentration of a pure solid or liquid is constant (i.e. independent of the amount present). In other words if a substance .X. is involved then [`X_((s))`and `X_((l))` ] are constant whatever the amount of .X. is taken. `therefore` Molarity =`"No. of MOL"/"VOLUME" xx "Density x Volume"/"Mass"` (`because` =`"Volume"/"Mass"`) and `"Density x Volume"/"Mass"` =1 `therefore` Molarity = Density At a given temperature density of pure solid and liquid is constant, So, concentration is constant. Any amount of solid and liquid taken but concentration during reaction is constant. So, pure solid and liquid can IGNORE. e.g.: `CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))` `K=([CaO_((s))][CO_2])/([CaCO_(3(s))])` So, `K_p=p_(CO_2)` and `K_c=[CO_2]`