Reactions involve breaking and making bonds

Chemical reactions involve the breaking and forming of bonds, which result in enthalpy changes.

The overall enthalpy change (ΔH) of a reaction is the combined effect of these two processes:

1. Breaking bonds in the reactant molecules - An endothermic process (positive ΔH) that requires energy. The breaking of stronger bonds requires more energy.
2. Forming bonds to create the product molecules - An exothermic process (negative ΔH) that releases energy. The formation of stronger bonds releases more energy.

The overall ΔH of a reaction depends on the relative amounts of energy involved in bond breaking and bond formation:

• If bond breaking requires more energy than is released during bond formation, the reaction has a positive ΔH.
• If bond breaking requires less energy than is released during bond formation, the reaction has a negative ΔH.

Bond energy

Bond energy (or bond enthalpy) is defined as the energy required to break one mole of a specified type of covalent bond in a gaseous molecule. Although a specific type of bond always requires a particular amount of energy to break, the exact amount can vary based on the bond's molecular environment.

For example, breaking each O-H bond in H_2_O involves different amounts of energy amounts. To account for these variations, chemists use average bond energies which represents the average energy needed to break a type of bond across various molecular environments.

Important points regarding bond enthalpy values include:

1. They are always positive (endothermic) because breaking bonds requires energy.
2. They can be exact or average values, with averages representing different molecular environments.

Calculating reaction enthalpy changes

The enthalpy change of a complete reaction (ΔH_r) is determined using average bond enthalpies for the bonds that are broken and those that are formed:

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

where Σ = sum of

Alternatively, ΔH_r can be expressed in terms of the bond enthalpies in the reactants and products:

ΔH_r = Σ(bond enthalpies in reactants) $-$ Σ(bond enthalpies in products)

Hess cycles provide more accurate ΔH_r values

Calculating ΔH_r using mean bond enthalpies is less accurate than using Hess cycles because average bond enthalpies do not account for the unique molecular environment of each bond in a specific molecule. Hess cycles, in contrast, consider the actual enthalpy changes associated with the specific bonds in the reactants and products, providing a more precise estimation of ΔH_r.

Worked example 1 - Calculating ΔH for the combustion of methane

Calculate the enthalpy change (ΔH) for the combustion of methane (CH₄) with oxygen to form carbon dioxide and water. The balanced chemical equation is:

CH₄ + 2O₂ ➔ CO₂ + 2H₂O

Given the mean bond enthalpies:

Step 1: Identify and count the bonds broken and formed

Bonds broken: 4 x C-H and 2 x O=O

Bonds formed: 2 x C=O and 4 x O-H

Step 2: Calculate total energy absorbed in breaking bonds

Energy to break bonds = (4 x 414) + (2 x 498) = 1,656 + 996 = 2,652 kJ mol^-1^

Step 3: Calculate total energy released in forming bonds

Energy released in bond formation = (2 x 799) + (4 x 464) = 1,598 + 1,856 = 3,454 kJ mol^-1^

Step 4: Calculate ΔH for the reaction

ΔH = Σbond enthalpies of bonds broken $-$ Σbond enthalpies of bonds formed

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

The negative sign indicates that the reaction is exothermic, releasing 802 kJ mol^-1^ of energy.

Worked example 2 - Calculating ΔH for the synthesis of ammonia

Calculate the enthalpy change (ΔH) for the synthesis of ammonia (NH₃) from nitrogen and hydrogen. The balanced chemical equation is:

N₂ + 3H₂ ➔ 2NH₃

Given the mean bond enthalpies:

Step 1: Identify and count the bonds broken and formed

Bonds broken: 1 x N≡N and 3 x H-H

Bonds formed: 6 x N-H

Step 2: Calculate total energy absorbed in breaking bonds

Energy to break bonds = (1 x 945) + (3 x 436) = 945 + 1,308 = 2,253 kJ mol^-1^

Step 3: Calculate total energy released in forming bonds

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

Step 4: Calculate ΔH for the reaction

ΔH = Σbond enthalpies of bonds broken $-$ Σbond enthalpies of bonds formed

ΔH = 2,253 $-$ 2,346 = $-$93 kJ mol^-1^

The negative sign indicates that the reaction is exothermic, releasing 93 kJ mol^-1^ of energy.