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

`XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) + F`
`Xe rarr Xe^(o+) +e^(-) (IE)`
`e^(-) +F rarr F^(Theta) (EA)`
`2F rarr F_(2) (Delta_(diss)H =- ve)`
`DeltaH = (4 xx 34) + (1 xx 279) +(1xx -38) +1 xx (-85)`
`= 292 kJ mol^(-1)`or
`XeF_(4)(s) rarr Xe +4F (DeltaH = 4 xx 34)`
`Xe rarr Xe^(o+) +e^(-) (IE = 279)`
`Fe +e^(-) rarr F^(Theta) (EA = - 85)`
`XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) +F`
`DeltaH = 4 xx 34 +279 - 85 - 38` ltbr. `= 292 kJ mol^(-1)`