Bond EnthalpyJEE Main

Concept

Breaking bonds costs energy; forming bonds releases it. A reaction's enthalpy is the difference: $\Delta H = \Sigma BE_{broken} - \Sigma BE_{formed}$. Methane burns exothermically because the strong C=O and O–H bonds formed release far more than the C–H and O=O bonds cost to break.

Key formula

Derivation

Imagine atomising all reactants (pay every bond enthalpy), then assembling products (collect every bond enthalpy). Hess's law guarantees the detour gives the true ΔH.

For CH₄ + 2O₂ → CO₂ + 2H₂O: broken = 4(C–H) + 2(O=O); formed = 2(C=O) + 4(O–H). With standard means: (1652 + 996) − (1598 + 1852) ≈ −802 kJ/mol.

Scenarios to explore

• Bond Enthalpy — ΔH from bonds broken minus bonds formed — CH₄ combustion.

Real-world applications

• Fuel comparison (per-gram energy of H₂ vs hydrocarbons).
• Estimating ΔH for reactions never run.
• Explaining why N₂ is inert (945 kJ/mol triple bond).

JEE exam tips

• BE method = gas-phase estimate; expect ±5% vs calorimetric values (mean-bond averaging).
• H₂O has 2 O–H bonds, CO₂ has 2 C=O — the two most-miscounted molecules.
• Stronger bond = shorter bond = higher BE (N≡N > N=N > N–N).

Common mistakes

• Formula backwards (products − reactants) — bond enthalpies are the reverse of formation-enthalpy logic.
• Counting bonds wrong — draw every structure fully.
• Using bond enthalpies for liquids/solids without vaporisation corrections.

Exam traps to avoid

• Bond enthalpy is always POSITIVE (breaking) — the sign lives in the formula.
• Resonance-stabilised products (CO₂!) release extra vs naive single-structure counts.