For the equilibrium : `CaCO_(3) (s) Leftrightarrow CaO(s)+CO_(2)(g)`
`triangleH_(f)^(0) (CaCO_(3))=-1207.1"kJ/mole", triangleH_(f)^(0) (CaO)=-635.5"kJ/mole"`
`triangleH_(f)^(0) (CO_(2))=-393.5"kJ/mole"`
(a) How would `K_p` depend on temperature
(b) The equilibrium constant for this reaction is much less than 1. Why, then does heating `CaCO_(3)(s)` in an open container lead to a complete conversion to the products?

Explanation

(a) `triangleH^(0)=[triangleH_(f)^(0) (CaO)+triangleH_(f)^(0) (CO_(2))]-triangleH_(f)^(0) (CaCO_(3))`
`=[-635.5+(-393.5)]-(-1207.1))`
=+178.1kJ
Because the given reaction is endothermic, the equilibrium constant, will increase with increasing temperature.
(a) On heating `CaCO_3, K_p` increases which favours the dissociation of `CaCO_3`. Also, in an open container, `CO_2` gas escapes and more `CaCO_3` dissociates to replace it unitil no more `CaCO_3` remains.