Calculate the enthalpy change (in kcal) for the reaction: . The average Xe – F bond enthalpy is 34 k — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate the enthalpy change (in kcal) for the reaction: $\text{XeF}_4 \to \text{Xe}^+ + \text{F}^- + \text{F}_2 + \text{F}$. 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 $\text{F}_2$ is 38 kcal/mol.
Answer: 0292
💡 Solution & Explanation
Break down the target reaction into standard stages: Atomization of $\text{XeF}_4$ ($4 \times 34 = 136\text{ kcal}$), Ionization of Xe ($279\text{ kcal}$), Electron gain for one F ($-85\text{ kcal}$), and reverse dissociation to form $\text{F}_2$ from two F atoms ($-38\text{ kcal}$). The fourth F atom remains as atomic F. Total $\Delta H = 136 + 279 - 85 - 38 = 292\text{ kcal}$. Format as 0292.
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