What is the bond enthalpy of Xe-F bond? XeF_(4)(g) rarr Xe^(+) (g) + F^(-)(g) + F_(2) (g) + F(g), Delta_(f)H=292kcal/mol. Given that I.E of Xenon =279k Cal/mole, B.E of F_(2)= 38 k Cal/mol. E.A of F=85k Cal/mole.

What is the bond enthalpy of Xe-F bond? XeF_(4)(g) rarr Xe^(+) (g) + F

1.	What is the bond enthalpy of Xe-F bond? XeF_(4)(g) rarr Xe^(+) (g) + F^(-)(g) + F_(2) (g) + F(g), Delta_(f)H=292kcal/mol. Given that I.E of Xenon =279k Cal/mole, B.E of F_(2)= 38 k Cal/mol. E.A of F=85k Cal/mole.
Answer» 24k Cal/mol 34 k Cal/mol 8.5 k Cal/mol 142.12 k Cal/mol Solution :Reactions for the given data are as follows `XeF_(4(g)) rarr Xe_((g))^(+) + F_((g))^(-) + F_(2(g)) + F_((g)) , Delta H_(1)` …(1) `Xe_((g)) rarr Xe^(+) + e^(-), Delta H_(2)` …(2) `F_(2(g)) rarr 2F_((g)) , Delta H_(3)` …(3) `F_((g)) + e^(-) rArr F_((g))^(-), Delta H_(4)` ...(4) From the above reactions we will giet the following reaction. `Xe_((g)) + 2F_(2(g)) rarr XeF_(4(g))`... (5) eq.(5) = eq.(2) + eq.(3) + eq(4)- eq.(1) `Delta H_(5) = 279 + 38- 85 - 292` `= -60 = 2` (B. E of `F_(2)`) - 4 (B.E of `Xe- F`) `=60 = 2(38) - 4(Xe - F)` (Xe-F) Bond energy = 34 K.Cal