13 public questions tagged with this topic.
Bond enthalpy increases with smaller halogen size due to stronger overlap. CH₃F has the highest C-X bond enthalpy.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nature of C-X Bond and Methods of Preparation
Bond enthalpy decreases as the halogen size increases due to weaker overlap. CH₃-I has the weakest C-X bond among F, Cl, Br, and I.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nature of C-X Bond and Methods of Preparation
For H₂(g) → 2H(g), Δ H = 2 × Δ Hf (H) - Δ Hf (H₂) = 2 × 218 - 0 = 436 kJ/mol, which is the bond enthalpy.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For CO(g) → C(g) + O(g) , bond enthalpy is Δ H = Δ Ha(C) + Δ Ha(O) - Δ Hf(CO) = 715.0 + 249.0 - (-110.5) = 1074.5 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For C₂H₆(g) → 2C(g) + 6H(g), Δ H = 2 × 715 + 6 × 218 - (-84.7) = 1430 + 1308 + 84.7 = 2822.7 kJ/mol. Bonds: 1(C-C) + 6(C-H), C-C = 2822.7 - 6 × 413 = 2822.7 - 2478 = 344.7 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
For H₂(g) + (1/2)O₂(g) → H₂O(g), Δ H = -241.8 kJ. Bonds broken: H-H (436) + (1/2)O=O. Bonds formed: 2(O-H) = 2 × 463 = 926. Δ H = (436 + (1/2)O=O) - 926 = -241.8, (1/2)O=O = 248.2, O=O = 496.4 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
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
For CCl₄(g) → C(g) + 4Cl(g) , Δ H = Δ Ha(C) + 2 × Δ Ha(Cl₂) - Δ Hf°(CCl₄,g) = 715 + 2 × 242 - (-95.7) = 715 + 484 + 95.7 = 1294.7 kJ/mol . Bond enthalpy = 1294.7 / 4 = 323.7 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
N₂ has a triple bond with a bond enthalpy of ~946 kJ/mol, higher than O₂ (498 kJ/mol), F₂ (159 kJ/mol), or Cl₂ (243 kJ/mol).
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Valence Bond Theory and Hybridisation - sp sp2 sp3 dsp2
HF has a single bond with a bond enthalpy of ~567 kJ/mol, higher than HCl (431 kJ/mol), HBr (366 kJ/mol), or HI (299 kJ/mol) due to strong electronegativity difference.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Kossel-Lewis Approach and Octet Rule and Lewis Structures
Cl₂ has a bond enthalpy of ~243 kJ/mol, higher than F₂ (159 kJ/mol), Br₂ (193 kJ/mol), or I₂ (151 kJ/mol) due to optimal bond strength.
Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds
