Calculate the enthalpy change (in kcal) for the reaction: XeF 6 → Xe + + F - + F 2 + 3F. The average Xe-F bond enthalpy is 32 kcal/mol, first ionization enthalpy of Xe is 278 kcal/mol, electron gain e

Question

Calculate the enthalpy change (in kcal) for the reaction: XeF 6 → Xe + + F - + F 2 + 3F. The average Xe-F bond enthalpy is 32 kcal/mol, first ionization enthalpy of Xe is 278 kcal/mol, electron gain enthalpy of fluorine is -84 kcal/mol and bond dissociation enthalpy of F 2 is 37 kcal/mol.

TestHub · Accepted Answer

For the reaction XeF 6 → Xe + 6F, the energy required is 6 × 32 = +192 kcal. The ionization of Xenon, Xe → Xe + , requires +278 kcal. The electron affinity of Fluorine, F + e - → F - , releases -84 kcal. The formation of a Fluorine molecule, 2F → F 2 , releases -37 kcal. Three Fluorine atoms remain unreacted. The total energy change is calculated as follows: 192 + 278 - 84 - 37 = 349 kcal Thus, the answer is 0349.

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