Chemical reactions involve breaking and forming bonds

In any chemical reaction, bonds between atoms in the reactants must be broken, and new bonds are formed to create the products.

• Breaking bonds requires an input of energy - This process is endothermic, meaning ΔH is positive.
• Forming bonds releases energy - This process is exothermic, meaning ΔH is negative.

The overall enthalpy change of a reaction (ΔH) depends on the difference between the energy required to break the bonds in the reactants and the energy released when new bonds form in the products:

• If more energy is needed to break the bonds than is released when forming new ones, ΔH will be positive, indicating an endothermic reaction.
• If less energy is required to break the bonds than is released when forming new ones, ΔH will be negative, indicating an exothermic reaction.

Bond enthalpy is the energy needed to break a bond

Bond enthalpy is the amount of energy required to break one mole of a particular type of bond in a molecule in the gas phase. For example, the O-H bond enthalpy is the energy needed to break one mole of O-H bonds in gaseous molecules such as water or alcohols.

Breaking a bond always requires energy input, so bond enthalpies are always positive values.

The energy needed to break a bond depends on the strength of attraction between the atoms involved:

• In ionic compounds, strong electrostatic forces of attraction exist between oppositely charged ions.
• In covalent molecules, the nuclei are attracted to the negatively charged electron pair that forms the covalent bond. Stronger bonds have higher bond enthalpies because more energy is required to overcome the forces of attraction between the atoms.

Bond enthalpies and reactivity

Bond enthalpies provide insight into how readily chemical bonds break during reactions. Bonds with high enthalpies require more energy to break, while those with lower enthalpies break more easily and are often the first to react.

For example, consider the reaction between hydrogen gas and chlorine gas to form hydrogen chloride:

H_2(g) + Cl_2(g) ➔ 2HCl_(g)

• The H-H bond has a relatively high enthalpy of 436 kJ mol^-1^, while the Cl-Cl bond has a lower enthalpy of 242 kJ mol^-1^.
• As a result, the Cl-Cl bond is more likely to break first in this reaction.

Reactions involving bonds with low enthalpies often occur readily at room temperature. In contrast, breaking high-enthalpy bonds typically requires additional energy input, such as heating or the use of a catalyst.

Mean bond enthalpies are average values

In reality, the actual bond enthalpy for a specific bond can vary slightly depending on the molecule in which it is found.

Mean bond enthalpies are defined as the energy needed to break one mole of bonds in the gas phase, averaged over many different compounds.

For instance, the two O-H bonds in a water molecule have different bond enthalpies:

• Breaking the first O-H bond requires 492 kJ mol^-1.
• Breaking the second O-H bond requires only 428 kJ mol^-1. This difference occurs because, after the first O-H bond is broken, the remaining O-H bond experiences greater electron repulsion, making it easier to break.

Mean bond enthalpies provided in data tables are average values taken from many different molecules containing that type of bond.

So, the O-H mean bond enthalpy of +463 kJ mol^-1 is the average of the bond enthalpies of the O-H bonds in molecules like water, alcohols, and carboxylic acids.

Calculating enthalpy changes using mean bond enthalpies

The overall enthalpy change of a reaction can be calculated using the following equation:

ΔH_reaction = Σ(bond enthalpies of bonds broken) $-$ Σ(bond enthalpies of bonds formed)

To use this method:

1. Identify all bonds broken in the reactants and all bonds formed in the products.
2. Sum the enthalpies of bonds broken (endothermic process, positive value).
3. Sum the enthalpies of bonds formed (exothermic process, negative value).
4. Subtract the sum of bonds formed from the sum of bonds broken.

Worked example 1 - Calculating the enthalpy change of a reaction

Calculate the enthalpy change for the reaction: CH_4(g) + 2O_2(g) ➔ CO_2(g) + 2H₂O_(g)

Given the following mean bond enthalpies:

Step 1: Identify and count the bonds broken and formed

Bonds broken: (4 × C-H) + (2 × O=O)

Bonds formed: (2 × C=O) + (4 × O-H)

Step 2: Calculate total energy absorbed in breaking bonds

Energy to break bonds = (4 × 414) + (2 × 498) = 2,652 kJ mol^-1

Step 3: Calculate total energy released in forming bonds

Energy released in bond formation = (2 × 799) + (4 × 464) = 3,454 kJ mol^-1

Step 4: Calculate ΔH for the reaction

ΔH_reaction = Σ(bond enthalpies of bonds broken) $-$ Σ(bond enthalpies of bonds formed)

ΔH_reaction = 2,652 $-$ 3,454 = $-$802 kJ mol^-1.

The negative value indicates that the reaction is exothermic.

There may be slight variations in the enthalpy change calculated from mean bond enthalpy data compared to an experimental value. This difference occurs because mean bond enthalpies are averaged values that do not account for small differences in actual bond enthalpies in specific molecules.

Calculating mean bond enthalpies from reaction enthalpies

When given the enthalpy change of a reaction and all but one of the bond enthalpies involved, you can calculate the unknown mean bond enthalpy by rearranging the equation:

ΔH_reaction = Σ(bond enthalpies of bonds broken) $-$ Σ(bond enthalpies of bonds formed)

Worked example 2 - Calculating mean bond enthalpy

For the reaction: N_2(g) + 3H_2(g) ➔ 2NH_3(g) ΔH_r, is $-$93 kJ mol^-1*.*Calculate the mean bond enthalpy of the N≡N bond.

Given the following mean bond enthalpies.

Step 1: Identify and count the bonds broken and formed

Bonds broken: (1 × N≡N) + (3 × H-H)

Bonds formed: (6 × N-H)

Step 2: Calculate total energy released in forming bonds

Energy released in bond formation = (6 × 391) = 2,346 kJ mol^-1

Step 3: Calculate total energy absorbed in breaking bonds

Σ(bond enthalpies of bonds broken) = ΔH_reaction + Σ(bond enthalpies of bonds formed)

Σ(bond enthalpies of bonds broken) = $-$93 + 2,346 = 2,253 kJ mol^-1

Step 4: Calculate mean bond enthalpy of N≡N bond

2,253 = (1 × N≡N) + (3 × 436)

N≡N = 2,253 $-$ 1,308 = +945 kJ mol^-1.

The mean bond enthalpy of the N≡N bond is 945 kJ mol^-1.