Bond enthalpies — estimating ΔH_rxn from bond energies

WHAT is a bond enthalpy?

HOW to derive the working formula (from first principles)

We want $\Delta H_{rxn}$. Enthalpy is a state function, so we can take any path between reactants and products. Choose the imaginary atomic path:

1. Path step 1 — atomize the reactants. Break every bond in the reactants → free gaseous atoms. Energy needed $= \sum (\text{bond enthalpies of reactant bonds}) = +\sum_{\text{reactants}} \Delta H_{bonds}$

2. Path step 2 — build the products. Form every bond in the products from those free atoms. Energy released $= -\sum (\text{bond enthalpies of product bonds}) = -\sum_{\text{products}} \Delta H_{bonds}$

Because enthalpy is path-independent (Hess's law), adding the two steps gives the reaction enthalpy:

Worked Example 1 — Hydrogenation of ethene

$\text{H}_2\text{C}\!=\!\text{CH}_2 + \text{H}_2 \longrightarrow \text{H}_3\text{C}\!-\!\text{CH}_3$

Bond values (kJ/mol): C=C 614, C–H 413, H–H 436, C–C 348.

Bonds broken (reactants):

• 1 × C=C = 614 — Why? The double bond in ethene must go.
• 4 × C–H = 1652 — Why? Ethene's 4 C–H bonds.
• 1 × H–H = 436 — Why? The added H₂.
• Total broken = $614 + 1652 + 436 = 2702$

Bonds formed (product ethane):

• 1 × C–C = 348 — Why? Double became single.
• 6 × C–H = 2478 — Why? Ethane has 6 C–H bonds.
• Total formed = $348 + 2478 = 2826$

$\Delta H_{rxn} = 2702 - 2826 = \mathbf{-124\ \text{kJ/mol}}$

Why negative? We destroyed a C=C + H–H and built a C–C + 2 extra C–H — net stronger bonding ⟹ exothermic. ✅ (matches forecast)

Worked Example 2 — Combustion of methane

$\text{CH}_4 + 2\,\text{O}_2 \longrightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}$

Values: C–H 413, O=O 498, C=O 799, O–H 463.

Broken: 4 C–H + 2 O=O $= 4(413) + 2(498) = 1652 + 996 = 2648$ Why 2 O=O? Coefficient 2 on O₂, each is one double bond.

Formed: 2 C=O (in CO₂) + 4 O–H (2 H₂O × 2 each) $= 2(799) + 4(463) = 1598 + 1852 = 3450$ Why 4 O–H? Two water molecules, each with two O–H bonds.

$\Delta H_{rxn} = 2648 - 3450 = \mathbf{-802\ \text{kJ/mol}}$

(Experimental ≈ −802 to −890; the gap is because table values are averages and H₂O here is gas, not liquid.)

WHY this is only an estimate

Active Recall

Connections

• Hess's Law — the state-function principle this derivation relies on.
• Standard Enthalpy of Formation — the more accurate alternative method.
• Enthalpy and the First Law of Thermodynamics
• Exothermic vs Endothermic Reactions
• Lewis Structures — needed to count bonds correctly.
• Resonance Energy — explains part of the estimation error.

Concept Map

Hinglish (regional understanding)