Calculate the enthalpy change for the following reaction:
`XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) +F`. The average `Xe-F` bond energy is `34 kcal mol^(-1)`, first `IE` of `Xe` is `279 kcal mol^(-1), EA` of `F` is `85 kcal mol^(-1)` and bond dissociation enegry of `F_(2)` is `38 kcal mol^(-1)`

Explanation

To calculate the enthalpy change for the reaction:

\[ \text{XeF}_4 \rightarrow \text{Xe}^{+} + \text{F}^{-} + \text{F}_2 + \text{F} \]

we will follow these steps:

### Step 1: Break down the reaction into individual processes
1. **Dissociation of XeF4 into Xe and F atoms**: This involves breaking 4 Xe-F bonds.
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