Following data is given for the reaction : `CaCO_3 (s) to CaO(s) +CO_2(g)`
Given that `Delta_fH^(@) [CaO(s)]=-635.1 "kJ mol"^(-1)`
`Delta_f H^@[CO_2(g)]=-393.5 "kJ mol"^(-1)`
`Delta_fH^@[CaCO_3(s)]=-1206.9 "kJ mol"^(-1)`
Predict the effect of temperature on the equilibrium constant of the above reaction.

Explanation

`delta_(r) H^(@) = Delta_(f) H^(@) [CaO (s) ] + Delta_(f) H^(@)[CO_(2)(g) ] -Delta_(f) H^(@) [CaCO_(3)(s)]`
`=- 635.1 + (-393.5) - (-1206.9) = 178.3 kJ "mol"^(-1)`
Thus, the reaction is endothermic. Hence, according to Le Chatelier's principle, on increasing the temperature, the equilibrium will proceed in the forward direction .