The average energy required to break a P-P bond in `P_(4)(s)` into gaseous atoms is `53.2 kcal mol^(-1)`. The bond dissociation energy of `H_(2)(g)` is `104.2kcal mol^(-1) , Delta H_(f)^(0)` of `PH_(3)(g)` from `P_(4)(s)` is `5.5 kcal mol^(-1)`. The P-H bond energy in kcal `mol^(-1)` is [ Neglect presence of Van der Waals force in `P_(4)(s)`]

Explanation

To find the P-H bond energy in kcal mol^(-1), we can follow these steps:

### Step 1: Write the reaction for the formation of PH3 from P4 and H2
The balanced reaction is:
\[ P_4(s) + 6H_2(g) \rightarrow 4PH_3(g) \]

### Step 2: Identify the energies involved
We know:
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