Concept Explanation: Calculating Bond Enthalpy of C=C Bond

We are given the enthalpy of formation from gaseous atoms for ethane (C₂H₆) and benzene (C₆H₆), along with the bond enthalpy of C-H and the resonance energy of benzene. We need to find the bond enthalpy of the C=C bond.

Key Idea: The enthalpy of formation from gaseous atoms (ΔH_f,atoms) is the energy released when bonds are formed to make the molecule from its individual atoms. For a molecule, it equals the negative sum of the bond enthalpies of all bonds in the molecule. However, for benzene, we must also account for its resonance energy, which provides extra stability.

Step-by-Step Solution

Step 1: Analyze the Molecule - Ethane (C₂H₆)

Ethane has 1 C-C bond and 6 C-H bonds.
Let the bond enthalpy of C-C be $E{C-C}_{}$.
Given bond enthalpy of C-H, $E{C-H}_{}=411.0\mathrm{kJ}/\mathrm{mol}$.
Given ΔH_f,atoms for ethane = -2839.2 kJ/mol.
The equation for the energy change is:
$-(1\times E{C-C}_{}+6\times E{C-H}_{})=\Delta H_{f,atoms}$
Substituting the values:
$-(E{C-C}_{}+6\times 411.0)=-2839.2$
Solving for $E{C-C}_{}$:
$E{C-C}_{}+2466.0=2839.2$
$E{C-C}_{}=2839.2-2466.0=373.2\mathrm{kJ}/\mathrm{mol}$

Step 2: Analyze the Molecule - Benzene (C₆H₆)

Benzene has a resonance structure. The standard Kekulé structure would have 3 C=C bonds, 3 C-C bonds, and 6 C-H bonds.
Let the bond enthalpy of C=C be $E{C=C}_{}$.
We already found $E{C-C}_{}=373.2\mathrm{kJ}/\mathrm{mol}$.
Given bond enthalpy of C-H, $E{C-H}_{}=411.0\mathrm{kJ}/\mathrm{mol}$.
Given ΔH_f,atoms for benzene = -5506 kJ/mol.
Given Resonance Energy (R.E.) = -23.68 kJ/mol (This is the extra stability; we must subtract it from the calculated bond energy sum).
The equation for the energy change, accounting for resonance, is:
$-\left[\right(3\times E{C=C}_{}+3\times E{C-C}_{}+6\times E{C-H}_{})-R.E.]=\Delta H_{f,atoms}$
Substituting the known values:
$-\left[\right(3\times E{C=C}_{}+3\times 373.2+6\times 411.0)-(-23.68\left)\right]=-5506$
Simplify inside the brackets:
$-[3E{C=C}_{}+1119.6+2466.0+23.68]=-5506$
$-[3E{C=C}_{}+3609.28]=-5506$
Multiply both sides by -1:
$3E{C=C}_{}+3609.28=5506$
Now, solve for $3E{C=C}_{}$:
$3E{C=C}_{}=5506-3609.28=1896.72$
Finally, solve for $E{C=C}_{}$:
$E{C=C}_{}=\frac{1896.72}{3}=632.24\mathrm{kJ}/\mathrm{mol}$

Final Answer: The bond enthalpy of the C=C bond is $632.24\mathrm{kJ}/\mathrm{mol}$.
This corresponds to the third option: 632.24 kJ/mol.

Related Topics & Formulae

1. Enthalpy of Formation from Gaseous Atoms (ΔH_f,atoms):
This is the enthalpy change when one mole of a compound is formed from its constituent atoms in the gaseous state. It is equal to the negative sum of the bond enthalpies of all the bonds in the molecule.
$\Delta H_{f,atoms}=\&$