Question:
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
Solution:
The given reaction CaO(s) + CO2(g) →CaCO3(s) is 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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