Question

Identify the molecule for which the enthalpy of atomization \( (\Delta_a H^\circ) \) and bond dissociation enthalpy \( (\Delta_{\text{bond}} H^\circ) \) are not equal.

• A. \( H_2 \)
• B. \( Cl_2 \)
• C. \( F_2 \)
• D. \( CH_4 \)

Hint:

For diatomic molecules like \( H_2, Cl_2, F_2 \), the enthalpy of atomization equals the bond dissociation enthalpy because there is only one bond. However, for polyatomic molecules like \( CH_4 \), different bond dissociation energies lead to unequal enthalpies.

Answer 1

The Correct Option is D

The enthalpy of atomization and bond dissociation enthalpy are not equal for CH₄. The key reason for this is that in CH₄, the process of atomization involves breaking bonds and forming isolated atoms, while bond dissociation only involves breaking specific bonds in a molecule.

For CH₄, the bond dissociation enthalpy refers to the energy required to break the C-H bonds in a single molecule, while atomization refers to the energy needed to break all the bonds and separate each atom in the molecule.

This difference is because in the atomization process, the breaking of bonds results in a much larger amount of energy release or absorption due to the total energy involved in transforming the entire molecule into isolated atoms. In contrast, bond dissociation involves less energy as only individual bonds are broken without altering the molecular structure significantly.