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

To determine which molecule has different enthalpy of atomization \( (\Delta_a H^\circ) \) and bond dissociation enthalpy \( (\Delta_{\text{bond}} H^\circ) \), let's consider the definitions and examine each option:

Enthalpy of Atomization: This is the enthalpy change when one mole of a compound in its standard state is converted into atoms.

Bond Dissociation Enthalpy: This is the enthalpy change required to break a particular bond in a molecule in its gaseous state.

For diatomic molecules like \( H_2 \), \( Cl_2 \), and \( F_2 \), the process of converting to atoms and breaking a bond is essentially the same, as they consist of two atoms bound by a single covalent bond.

However, in the case of \( CH_4 \) (methane), the molecule consists of one carbon atom bonded to four hydrogen atoms:

• \( CH_4 \rightarrow C + 4H \)

Here, the enthalpy of atomization involves breaking four C-H bonds to convert one mole of methane to its constituent atoms, which is more complex than a single bond dissociation.

Thus, the enthalpy of atomization for \( CH_4 \) will not be equal to the bond dissociation enthalpy of a single C-H bond, unlike diatomic molecules where both values would be equal.

Therefore, the molecule for which \( \Delta_a H^\circ \) and \( \Delta_{\text{bond}} H^\circ \) are not equal is \( CH_4 \).

Answer 2

The enthalpy of atomization and bond dissociation enthalpy are not equal for CH₄. 

The key reason is that in CH₄, the process of atomization involves breaking all bonds in the molecule to form individual atoms, whereas bond dissociation enthalpy refers to the energy required to break a specific bond in the molecule.

Bond dissociation enthalpy of CH₄: Refers to the energy needed to break one C–H bond in a methane molecule, typically resulting in the formation of CH₃ and H radicals.

Enthalpy of atomization of CH₄: Refers to the total energy required to break all four C–H bonds to form individual atoms (1 C and 4 H atoms) in the gaseous state.

This difference arises because atomization is a complete breakdown of the molecular structure into atoms, which requires more energy than simply breaking one bond. Thus, the enthalpy of atomization is generally higher than the bond dissociation enthalpy for a single bond in CH₄.