Following data is given for the reaction: CaCO_(3(s)) to CaO_((s)) + CO_(2(g)) Delta_f H^ө [CaO_((s))]=-"635.1 kJ mol"^(-1) Delta_f H^ө [CO_(2(g))]=-393.5 "kJ mol"^(-1) Delta_f H^ө [CaCO_(3(s))]=-1206.9 "kJ mol"^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.

Following data is given for the reaction: CaCO_(3(s)) to CaO_((s)) + C

1.	Following data is given for the reaction: CaCO_(3(s)) to CaO_((s)) + CO_(2(g)) Delta_f H^ө [CaO_((s))]=-"635.1 kJ mol"^(-1) Delta_f H^ө [CO_(2(g))]=-393.5 "kJ mol"^(-1) Delta_f H^ө [CaCO_(3(s))]=-1206.9 "kJ mol"^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.
Answer» Solution :GIVEN that, `Delta_f H^ө [CaO_((s))]=-"635.1 kJ mol"^(-1)` `Delta_f H^ө [CO_(2(g))]=-393.5 "kJ mol"^(-1)` `Delta_f H^ө [CaCO_(3(s))]=-1206.9 "kJ mol"^(-1)` In the REACTION `CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))` `Delta_f H^ө=Delta_f H^ө [CaO_((s))]+ Delta_f H^ө [CO_(2(g))]-Delta_f H^ө [CaCO_(3(s))]` `THEREFORE Delta_f H^ө`=-635.1+(-393.5)-(-1206.9) =178.3 kJ `"mol"^(-1)` As `DeltaH` value is positive, hence the reaction is ENDOTHERMIC. So, according to Le-Chatelier.s principle, reaction will proceed in FORWARD direction on increasing temperature.