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प्रश्न
Standard molar enthalpy of formation, ∆fHΘ is just a special case of enthalpy of reaction, ∆rHΘ. Is the ∆rHΘ for the following reaction same as ∆fHΘ? Give reason for your answer.
\[\ce{CaO(s) + CO2(g) -> CaCO3(s); ∆_fH^Θ = - 178.3 kJ mol^{-1}}\]
उत्तर
No, standard molar enthalpy of formation is the enthalpy of the reaction when 1 mole of compound is formed from its constituent elements.
\[\ce{Ca(s) + C(s) + 3/2 O2(g) -> CaCO3(s); Δ_fH^Θ}\]
As the close reaction is different from the given reaction,
∴ \[\ce{Δ_rH^Θ ≠ Δ_fH^Θ}\]
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