Q37 of 54 Page 90

Following data is given for the reaction: CaCO3 (s) → CaO3 (s) + CO2 (g)

ΔfH⊖ [CaO (s)] = -635.1 kJ mol=-1


ΔfH⊖ [CO2 (g)] = -393.5 kJ mol-1


ΔfH⊖ [CaCO3 (s)] = -1206.9 kJ mol-1


Predict the effect of temperature on the Equilibrium constant of the above reaction.


Given; CaCO3 (s) → CaO3 (s) + CO2 (g)

ΔfH⊖ [CaO (s)] = -635.1 kJ mol-1


ΔfH⊖ [CO2 (g)] = -393.5 kJ mol-1


ΔfH⊖ [CaCO3 (s)] = -1206.9 kJ mol-1


We know that,



Where is the standard enthalpy change for the formation of one mole of compound from its elements in their most stable states of aggregation.


= ΔfH⊖ [CaO(s)] + ΔfH⊖ [CO2(g)]- ΔfH⊖ [CaCO3(s)]


= -635.1 -393.5+1206.9


=178.3 kJ mol-1


=positive


i.e. the reaction is exothermic.


Hence according to Le Chatelier’s Principle on increasing the temperature the reaction will shift to forward direction.


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Equilibrium constant Kc =


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Column I (Reaction) Column II (Equilibrium constant)