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Δ H = Δ Hf° (C₂H₆) - [Δ Hf° (C₂H₂) + 2 × Δ Hf° (H₂)] = -84.7 - [226.7 + 0] = -84.7 - 226.7 = -311.4 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
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
The reaction forms 1 mol of C₂H₄(g) from its elements in standard states, so Δ H = Δ Hf° (C₂H₄) = 52.3 kJ/mol, as Δ Hf° (C) = 0 and Δ Hf° (H₂) = 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
Δ H = 2 × Δ Hf° (NH₃) - [Δ Hf° (N₂) + 3 × Δ Hf° (H₂)] = 2 × (-46.1) - [0 + 0] = -92.2 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
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