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Question: What is the bond enthalpy of \(Xe - F\) bond? \(Xe{F_4} \to X{e^ + }(g) + {F^ - }(g) + {F_2}(g) + ...

What is the bond enthalpy of Xe−FXe - F bond?
XeF4→Xe+(g)+F−(g)+F2(g)+F(g); ΔHr=292kcal/molXe{F_4} \to X{e^ + }(g) + {F^ - }(g) + {F_2}(g) + F(g);{\text{ }}\Delta {H_r} = 292kcal/mol
Given: Ionization for Xe =279kcal/mol = 279kcal/mol , B.E(F−F)=38kcal/molB.{E_{(F - F)}} = 38kcal/mol , Electron Affinity of F =85kcal/mol = 85kcal/mol
A) 24 kcal/mol
B) 34 kcal/mol
C) 8.5 kcal/mol
D) None of these

Explanation

Solution

The enthalpy for a reaction is denoted by ΔrH{\Delta _r}H and is given as the difference of enthalpies between product and reactants. The standard molar enthalpies can be given as ΔrH0{\Delta _r}{H^0} . The heat that is absorbed or released by the reaction is known as it’s enthalpy.

Complete answer:
In the given question the ΔrH{\Delta _r}H is already given. The ΔrH{\Delta _r}H would be equal to the total or sum of all the enthalpies of the reaction taking place in the process.
The sub-reactions that are occurring are:
Breaking of the Xe-F bond denoted by eXe−F{e_{Xe - F}} . Since there are 4 Xe-F bonds we will multiply eXe−F{e_{Xe - F}} by four.
Conversion of Xe→Xe+;IE1(Xe)=279kcal/molXe \to X{e^ + };I{E_1}(Xe) = 279kcal/mol
Conversion of F→F−;EA1(F2)=−85kcal/molF \to {F^ - };E{A_1}({F_2}) = - 85kcal/mol (since the electron gain enthalpy is always negative)
We have been given the bond breaking energy of fluorine B.E(F−F)=38kcal/molB.{E_{(F - F)}} = 38kcal/mol , but in the given reaction the bond is forming between the two Fluorine atoms. Bond forming will be the inverse of Bond Breaking. The reaction can be given as:
F+F→F2;eF2=−38kcal/molF + F \to {F_2};{e_{{F_2}}} = - 38kcal/mol
The total enthalpy of the reaction is the sum of all these enthalpies. Mathematically it can be expressed as:
ΔrH=IE1(Xe)+EA1(F2)+eF2+4×eXe−F{\Delta _r}H = I{E_1}(Xe) + E{A_1}({F_2}) + {e_{{F_2}}} + 4 \times {e_{Xe - F}}
Substituting the respective values, we get eXe−F{e_{Xe - F}} as:
4×eXe−F=ΔrH−IE1(Xe)−EA1(F2)−eF24 \times {e_{Xe - F}} = {\Delta _r}H - I{E_1}(Xe) - E{A_1}({F_2}) - {e_{{F_2}}}
eXe−F=292−279−(−85)−(−38)4=34kcal/mol{e_{Xe - F}} = \dfrac{{292 - 279 - ( - 85) - ( - 38)}}{4} = 34kcal/mol

Hence the correct answer is Option (B).

Note:
Remember that electron gain enthalpy is always negative as energy is released for gaining an electron and Ionisation enthalpy is always positive as energy is absorbed with the loss of the electron. For finding the total enthalpies of a reaction always the Hess Law is used, which states that the enthalpy of a reaction is the sum of the enthalpies of the constituent reactions.