Standard molar enthalpy of formation, Δf HΘis just a special case of enthalpy of reaction, Δr HΘ. Is the Δr HΘfor the following reaction same as Δf HΘ? Give the reason for your answer.

Cao(s) + CO2(g) CaCO3(s) ; ΔfHΘ = –178.3 kJ mol–1



o The Heat of Reaction or the enthalpy of a reaction Δr HΘ is the change in the enthalpy of a chemical reaction that occurs at a constant pressure.


Whereas, the standard molar enthalpy of formation ΔfHΘ is defined as the change in enthalpy when one mole of substances are involved in the standard state (1 atm of pressure and 298.15 K) formed from pure elements under the same standard conditions; hence it is a special case of enthalpy of reaction (in addition with the 1 mole and standard forms).


o The given reaction CaO(s) + CO2(g) CaCO3(s) is clearly indicating that it is occurring in the standard form of 1 mole of each substance. And the molar enthalpy of formation ΔfHΘ = –178.3 kJ mol–1 Given for CaCO3 is also showing the standard conditions.


So,ΔfHΘ = –178.3 kJ mol–1 = Δr HΘ .


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