Chemistry - Thermochemistry Question with Solution | TestHub
Calculate the enthalpy change (in kcal) for the reaction: XeF4 → Xe+ + F- + F2 + F. The average Xe-F bond enthalpy is 34 kcal/mol, first ionization enthalpy of Xe is 279 kcal/mol, electron gain enthalpy of fluorine is -85 kcal/mol and bond dissociation enthalpy of F2 is 38 kcal/mol.
Answer:
Solution:
To determine the total energy change, we consider the following steps:
1. **Breaking Xe–F bonds in XeF₄:**
Breaking 4 Xe–F bonds requires 4 × 34 = +136 kcal.
2. **Ionization of Xe:**
The ionization of Xe to Xe⁺ requires +279 kcal.
3. **Electron affinity of F:**
The electron affinity of F, where F + e⁻ → F⁻, releases -85 kcal.
4. **Formation of F₂:**
The formation of F₂ from 2F atoms releases -38 kcal. Note that one F atom remains as an atom.
The total energy change is the sum of these values:
136 + 279 - 85 - 38 = 292 kcal
Therefore, the answer is 0292.
