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Δ H = Δ Hf° (C₂H₆) - [2 × Δ Hf° (C) + 3 × Δ Hf° (H₂)] = -84.7 - [0 + 0] = -84.7 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
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