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.

Average bond enthalpies

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 enthalpy, which represents the average energy needed to break one mole a type of covalent bonds in gaseous molecules.

Important points regarding bond enthalpy values include:

1. They are always positive (endothermic) because breaking bonds requires energy.
2. They represent an average value across 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.