Enthalpy of Atomization

Enthalpy of Atomization refers to the amount of energy required to break all the bonds in one mole of a substance to convert it into individual atoms in the gas phase. It is typically expressed in kilojoules per mole (kJ/mol) and can be understood as the energy needed to "atomize" a substance into its component atoms.

1.0Definition of Enthalpy of Atomization: 

The enthalpy of atomization (ΔH_atom​) is the energy change when one mole of a compound, in its standard state, is converted into individual gaseous atoms. This process involves breaking the chemical bonds between atoms within a molecule or crystal lattice.

                    Compound (solid/liquid/gas) ⟶ Atoms

For example, in the case of diatomic hydrogen (H₂), the enthalpy of atomization is the energy required to break the bond between two hydrogen atoms to form two separate hydrogen atoms in the gas phase:

                      H₂(g) ⟶ 2H(g)   ΔaH⁰= 435.0 kJ mol^-1

2.0Factors Affecting Enthalpy of Atomization:

• Nature of Bonds: 

The type of bonding (metallic, covalent, or ionic) greatly influences the enthalpy of atomization. Stronger bonds require more energy to break, resulting in higher enthalpy values.

• Molecular Structure: 

The molecular structure and the number of bonds per atom in a molecule also affect the enthalpy of atomization. Compounds with a more compact and stable atomic structure typically have higher enthalpies.

3.0Solved Questions

Q. 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.

Ans. 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

Q. If Enthalpy of atomization of C=a, Bond enthalpy of H₂=b, Enthalpy of formation of CH₄=c, Enthalpy of formation of  C₂H₆=d. What is the C−C bond energy?

Ans. 

Given:

• Enthalpy of atomization of C=a
• Bond enthalpy of H₂=b
• Enthalpy of formation of CH₄=c
• Enthalpy of formation of C₂H₆=d

Steps to Calculate C−C Bond Energy:

Methane Formation: C(g) + 2H₂(g) → CH₄(g)

The enthalpy change for forming methane is: a + 2b = c $\to$b = (c−a)/2

Ethane Formation: 2C(g) + 3H₂(g) → C₂H₆(g)

The enthalpy change for forming ethane is: 2a + 3b − E_C−C = d

For E_C−C

                                       E_C−C =  2a + 3 (c-a)/2 − d

                                      E_C−C = a/2− d + 3c/2