6 public questions tagged with this topic.
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 = 2 × Δ Hf° (NO₂) - [Δ Hf° (N₂) + 2 × Δ Hf° (O₂)] = 2 × 33.2 - [0 + 0] = 66.4 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
The standard enthalpy of formation is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states, i.e., C(s) + O₂(g) → CO₂(g).
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law
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
For N₂(g) + 3H₂(g) → 2NH₃(g) , Δ H = 2 × (-46.10) = -92.20 kJ . Bonds broken: N≡N + 3 × H-H = N≡N + 3 × 436.0 = N≡N + 1308.0 kJ . Bonds formed: 6 × N-H = 6 × 391.0 = 2346.0 kJ . Using Δ H = (bonds broken) - (bonds formed) , -92.20 = (N≡N + 1308.0) - 2346.0 , N≡N = 945.8 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For O₂(g) → 2O(g), Δ H = 498 kJ/mol. Δ Hf (O₂) = 0, so Δ H = 2 × Δ Hf (O) - 0, Δ Hf (O) = 498 / 2 = 249 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
