6.2 Bond Energy

When a chemical reaction occurs, chemical bonds in the reactant molecules are broken, and new bonds are formed in the product molecules. This process involves changes in energy.

Bond Breaking: This process always requires an input of energy. The energy needed to break one mole of a particular bond to form neutral gaseous atoms is known as the bond dissociation energy. It is an endothermic process ($\Delta H=+ve$).

Bond Formation: This process always releases energy. The energy released when one mole of a bond is formed from neutral gaseous atoms is called bond energy. It is an exothermic process ($\Delta H=-ve$).

Calculating Enthalpy Change from Bond Energies

The overall energy change in a reaction, known as the enthalpy change ($\Delta H$), is the difference between the energy absorbed to break bonds and the energy released when new bonds are formed.

$\Delta H=\sum H_{\text{(bonds broken)}}-\sum H_{\text{(bonds formed)}}$

Where:

• $\sum H_{\text{(bonds broken)}}$ = sum of bond dissociation energies of all bonds in the reactants
• $\sum H_{\text{(bonds formed)}}$ = sum of bond energies of all bonds in the products

Worked Examples

Example 1: Theoretical Bond Energy of H–Cl

Calculate the bond energy for the H–Cl bond using the reaction:

${\mathrm{H}}_2(\mathrm{g})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})\to 2\mathrm{HCl}(\mathrm{g})$

Given:

• Bond dissociation energy of H–H: $436\text{kJ/mol}$
• Bond dissociation energy of Cl–Cl: $243\text{kJ/mol}$

Solution:

1. Bonds broken (reactants): 1 mol H–H + 1 mol Cl–Cl
2. Bonds formed (products): 2 mol H–Cl
3. Energy balance:

$436+243=2\times (\text{Bond energy of H–Cl})$

$679\text{kJ}=2\times (\text{Bond energy of H–Cl})$

$\text{Bond energy of H–Cl}=\frac{679}{2}=339.5\text{kJ/mol}$

Example 2: Enthalpy of Reaction for Water Formation

Calculate $\Delta H$ for:

$2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to 2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Given:

• H–H bond energy: $436\text{kJ/mol}$
• O=O bond energy: $493.6\text{kJ/mol}$
• O–H bond energy: $460\text{kJ/mol}$

Solution:

$\Delta H=1365.6-1840=-474.4\text{kJ}$

The negative sign confirms the reaction is exothermic, releasing $474.4\text{kJ}$ of energy.

Example 3: Average Bond Energy of C–H in Methane

Calculate the average bond energy per mole of C–H bonds in methane ($\mathrm{C}{\mathrm{H}}_4$).

Given: $\mathrm{C}{\mathrm{H}}_4(\mathrm{g})\to \mathrm{C}(\mathrm{s})+4\mathrm{H}(\mathrm{g})\Delta H^{\circ }=+1662\text{kJ}$

Solution:

One molecule of $\mathrm{C}{\mathrm{H}}_4$ contains four C–H bonds. The total energy absorbed ($+1662\text{kJ}$) breaks all four bonds.

$\text{Average C–H bond energy}=\frac{1662\text{kJ}}{4}=+415.5\text{kJ/mol}$

6.2.1 Variability in Bond Energies

Tabulated bond energy values are often averages, and their precision depends on the specific molecule and its chemical environment.

Well-Defined (Exact) Bond Energies

For diatomic homonuclear molecules (e.g., ${\mathrm{H}}_2,{\mathrm{O}}_2,{\mathrm{N}}_2,\mathrm{C}{\mathrm{l}}_2$), bond energies are highly accurate and precise. Since only one type of bond exists in the molecule, the bond dissociation energy is the same for every molecule of that type.

Approximate Bond Energies

In polyatomic molecules, bond energies are approximate. The strength of a particular bond (e.g., C–H or C–C) is influenced by:

• Neighboring atoms: Other atoms or functional groups attached to the bonded atoms alter the electron density and thus the bond strength. For example, the C–H bond energy in $\mathrm{C}{\mathrm{H}}_4$ differs slightly from that in $\mathrm{CHC}{\mathrm{l}}_3$.
• Molecular environment: Resonance, inductive effects, and hybridization all affect bond strength.

Because of this variability, bond energies for bonds in polyatomic molecules (e.g., C–H, C–C, C=O) are average values calculated from many different compounds. These averages are useful for estimating $\Delta H$ but may differ from experimentally measured values.

6.2.2 Factors Affecting Bond Energy

The following factors determine the strength of a covalent bond:

1. Bond Length

Bond energy is inversely proportional to bond length. Shorter bonds have greater orbital overlap, resulting in stronger bonds and higher bond energy.

2. Bond Multiplicity (Bond Order)

Bond energy increases with bond order: $C\equiv C(837\text{kJ/mol})>C=C(614\text{kJ/mol})>C-C(347\text{kJ/mol})$

Higher bond order also means shorter bond length.

3. Bond Polarity

Greater electronegativity difference between bonded atoms increases ionic character, which adds electrostatic attraction and generally increases bond energy. This explains why HF ($569\text{kJ/mol}$) has a higher bond energy than HI ($297\text{kJ/mol}$).

4. Lone Pair Repulsion (Anomalous Cases)

In small atoms like fluorine, lone pairs on adjacent atoms are forced very close together, causing lone pair–lone pair repulsion that weakens the bond. This explains why $F_2$ ($158\text{kJ/mol}$) has a lower bond energy than $C{l}_2$ ($242\text{kJ/mol}$) despite fluorine's higher electronegativity.

Using Bond Energy to Compare Reactivity

Bond energy values allow us to compare the reactivity of covalent molecules:

• Molecules with lower bond energy are more reactive (bonds break more easily)
• Molecules with higher bond energy are less reactive (bonds are harder to break)

Example: $F_2$ is more reactive than $C{l}_2$ not because of bond energy alone, but $I_2$ is less reactive than $F_2$ in terms of bond strength considerations alongside other factors.

Among hydrogen halides: HI is most reactive (lowest bond energy, $297\text{kJ/mol}$); HF is least reactive (highest bond energy, $569\text{kJ/mol}$).