Standard molar enthalpy of formation, Δ f H Θ is just a special case of enthalpy of reaction, Δ r H Θ . Is the Δ r H Θ for the following reaction same as Δ f H Θ ? Give the reason for your answer. Cao(s) + CO 2 (g) → CaCO 3 (s) ; Δ f H Θ = –178.3 kJ mol –1

Question

Philoid · Accepted Answer

o The Heat of Reaction or the enthalpy of a reaction Δ_r H^Θ is the change in the enthalpy of a chemical reaction that occurs at a constant pressure.

Whereas, the standard molar enthalpy of formation Δ_fH^Θ 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 Δ_fH^Θ = –178.3 kJ mol^–1 Given for CaCO₃ is also showing the standard conditions.

So,Δ_fH^Θ = –178.3 kJ mol^–1 = Δ_r H^Θ .

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

Standard molar enthalpy of formation, Δ_f H^Θis just a special case of enthalpy of reaction, Δ_r H^Θ. Is the Δ_r H^Θfor the following reaction same as Δ_f H^Θ? Give the reason for your answer.
Cao(s) + CO₂(g) → CaCO₃(s) ; Δ_fH^Θ = –178.3 kJ mol^–1