The `C-H` bond of the side chain in toluene, `C_(6)H_(5)-CH_(3)`, has a dissociation energy of `77.5 kcal mol^(-1)`. Calculate `Delta_(f)H^(Theta)` of benzy`1` radical and the strength of the central bond in dibenzy`1 C_(6)H_(5)-CH_(2)-CH_(2)-C_(6)H_(5)` given that `Delta_(f)H^(Theta)` to toluene vapour in `12 kcal mol^(-1)` and that of dibenzy`1` vapour is `27.8 kcal mol^(-1). BE` of `H_(2) = 104 kcal mol^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Enthalpy of Formation of Benzyl Radical

We know the dissociation energy of the C-H bond in toluene (C₆H₅-CH₃) is given as 77.5 kcal/mol. The reaction can be represented as:

\[ \text{C}_6\text{H}_5\text{CH}_3 \rightarrow \text{C}_6\text{H}_5\text{CH}_2 + \text{H} \]

...