Bond energy calculation — find ΔH of reaction from bond dissociation energies

Question

Calculate the enthalpy of the reaction $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$ using bond dissociation energies. Given: C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 799 kJ/mol, O-H = 463 kJ/mol.

(NCERT Class 11, Chapter 6)

Solution — Step by Step

In $\text{CH}_4$ : 4 C-H bonds broken

In $2\text{O}_2$ : 2 O=O bonds broken

Energy absorbed (bonds broken):

\Delta H_{broken} = 4 \times 413 + 2 \times 498 = 1652 + 996 = 2648 \text{ kJ}

In $\text{CO}_2$ : 2 C=O bonds formed

In $2\text{H}_2\text{O}$ : $2 \times 2 = 4$ O-H bonds formed

Energy released (bonds formed):

\Delta H_{formed} = 2 \times 799 + 4 \times 463 = 1598 + 1852 = 3450 \text{ kJ}

\Delta H_{rxn} = \text{Energy absorbed} - \text{Energy released}

\Delta H_{rxn} = \Delta H_{broken} - \Delta H_{formed}

\Delta H_{rxn} = 2648 - 3450

\boxed{\Delta H_{rxn} = -802 \text{ kJ/mol}}

The negative sign confirms this is an exothermic reaction — consistent with combustion of methane.

Why This Works

Bond breaking requires energy (endothermic), while bond formation releases energy (exothermic). The enthalpy of a reaction is the net balance: if more energy is released in forming new bonds than is consumed in breaking old bonds, the reaction is exothermic.

The formula $\Delta H = \Sigma(\text{BDE of bonds broken}) - \Sigma(\text{BDE of bonds formed})$ works because we are imagining a hypothetical path: first break all bonds in the reactants (go to atoms), then form all bonds in the products (come back from atoms). By Hess’s law, the enthalpy change depends only on the initial and final states, not the path.

Alternative Method

You can also use standard enthalpies of formation ( $\Delta H_f°$ ) if available:

\Delta H_{rxn} = \Sigma \Delta H_f°(\text{products}) - \Sigma \Delta H_f°(\text{reactants})

This gives a more accurate result because bond energies are averages and can vary slightly depending on the molecular environment.

For JEE, memorise these common bond energies: C-H (413), C-C (348), C=C (614), C=O (799), O-H (463), O=O (498), N-H (391), H-H (436). With these, you can handle almost every bond energy numerical that appears.

Common Mistake

The most frequent error is reversing the formula — writing $\Delta H = \text{bonds formed} - \text{bonds broken}$ instead of $\text{bonds broken} - \text{bonds formed}$ . Remember: breaking bonds costs energy (positive), forming bonds releases energy (negative). The formula is $\Delta H = \Sigma BDE_{\text{reactants}} - \Sigma BDE_{\text{products}}$ . If you get a positive value for combustion, you have flipped the sign.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next