Bond Enthalpy

Bond enthalpy, also known as bond energy, measures the strength of a chemical bond by quantifying the energy required to break 1 mole of a specific bond. For example, the bond enthalpy of an oxygen-hydrogen single bond is 463 kJ/mol, meaning 463 kJ of energy is needed to break 1 mole of such bonds.

1.0Energy Associated with Chemical Bonds

Chemical reactions involve breaking and forming chemical bonds, and energy plays an important role in these processes. Energy is required to break bonds (endothermic process), and energy is released when bonds are formed (exothermic process). The total heat change in a chemical reaction can be related to the energy changes associated with these bond-breaking and bond-forming processes.

In thermodynamics, we use two important terms to describe the energy associated with chemical bonds:

1. Bond dissociation enthalpy
2. Mean bond enthalpy

Let’s break these down and discuss their significance with reference to diatomic and polyatomic molecules.

Bond Dissociation Enthalpy

The bond dissociation enthalpy refers to the amount of energy required to break one mole of a specific covalent bond in a gaseous molecule, resulting in the formation of gaseous atoms. It is essentially the enthalpy change when one mole of bonds is broken.

Diatomic Molecules:

The bond dissociation enthalpy is straightforward for diatomic molecules (which consist of two atoms). For example, breaking the bond in one mole of hydrogen gas (H₂) requires an enthalpy change:

H₂(g) → 2H(g)  ΔH_dissociation = 435.0 kJ/mol

In this case, 435.0 kJ/mol is the bond dissociation enthalpy of the H-H bond. This process is the same as the enthalpy of atomization for hydrogen gas because breaking all bonds in diatomic molecules results in free atoms.

Other examples include:

• Chlorine (Cl₂): Cl₂(g) → 2Cl(g)  ΔH_dissociation = 242 kJ/mol
• Oxygen (O₂): O₂(g) → 2O(g)   ΔH_dissociation = 428 kJ/mol

Mean Bond Enthalpy

For polyatomic molecules (which consist of more than two atoms), different bonds within the molecule may require different amounts of energy to break. For example, in methane (CH₄), each C-H bond is not identical in energy when broken individually, despite their similar bond lengths and overall behavior.

When discussing polyatomic molecules, we use the mean bond enthalpy, which is the average energy needed to break a particular type of bond in a molecule. In this context, we average the bond dissociation energies for each successive bond-breaking step.

Polyatomic Molecules:

Take methane (CH₄) as an example. The overall thermochemical equation for its atomization is:

CH₄(g) → C(g) + 4H(g)   ΔH_atomization = 1665 kJ/mol

However, breaking each C-H bond one at a time requires a slightly different amount of energy:

• Breaking the first C-H bond: 

CH₄(g) → CH₃(g) + H(g)    ΔH = 427 kJ/mol

• Breaking the second C-H bond: 

CH₃(g) → CH₂(g) + H(g)   ΔH = 439 kJ/mol

• Breaking the third C-H bond: 

CH₂(g) → CH(g) + H(g)  ΔH=452 kJ/mol

• Breaking the fourth C-H bond: 

CH(g)→C(g) + H(g)  ΔH = 347 kJ/mol

The mean bond enthalpy is calculated by averaging the energy required to break all the C-H bonds:

Mean C-H bond enthalpy = 1665 kJ/mol

This shows that the mean C-H bond enthalpy in methane is 416 kJ/mol, though the exact values for each bond-breaking step differ slightly.

2.0Calculating Reaction Enthalpy Using Bond Enthalpies

The standard enthalpy of a reaction (ΔH) in the gas phase can be estimated using the bond enthalpies of the reactants and products. This is particularly useful when the enthalpies of formation are not readily available. The relationship is given by:

ΔH_reaction = ∑Bond enthalpies of reactants − ∑Bond enthalpies of products

This means that to calculate the reaction enthalpy, we:

1. Sum the bond enthalpies required to break all bonds in the reactant molecules.
2. Subtract the bond enthalpies of the bonds formed in the product molecules.

Example:

Consider the reaction:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

To calculate the reaction enthalpy, we use bond enthalpies for the following bonds:

Reactants:

• 4 C-H bonds in CH₄
• 2 O=O bonds in O₂

Products:

• 2 C=O bonds in CO₂
• 4 O-H bonds in H₂O

By using known bond enthalpy values, we can estimate the overall enthalpy change for the reaction.

3.0Solved Example of Bond Enthalpy

1. If E_C–C is 344 kJ mol^–1 and E_C–H is 415 kJ mol^–1, calculate the enthalpy of formation of propane. The enthalpies of atomization of carbon(s) and hydrogen (g) are 716 kJ mole^–1 and 433 kJ mole^–1 respectively.

Solution:

The enthalpy of formation is the sum of the atomization and bond energies. For propane, the enthalpies of atomization are

     3C(s) → 3C(g) ; ΔH = 3 × 716 = 2148 kJ

     4H₂(g) → 8H(g) ; ΔH = 4 × 433 = 1732 kJ

The bond enthalpies are

     2E_C–C = 2 × –344 = –688 kJ

     8E_C–H = 8 × –415 = –3320 kJ

Adding

     3C + 4H₂ → C₃H₈ ; ΔH_f = 2148 + 1732 – 688 – 3320 = –128 kJ mole^–1

2. The enthalpy changes at 298 K in successive breaking of O – H bonds of H – O – H are

H₂O_(g) → H_(g) + OH_(g),   ΔH = 498 kJ mol^–1

OH_(g) → H_(g) + O_(g),   ΔH = 428 kJ mol^–1

The bond enthalpy of the O – H bond is-

Solution:

The total enthalpy change for breaking both O–H bonds in H₂O is: 

498 kJ mol⁻¹+428 kJ mol⁻¹= 926 kJ mol⁻¹

(B.E.)_av =  926 / 2 = 463 kJ