Respuesta :
Xe +f2 →Xef2
ΔXe = ΣB.P reactants - Σ B.d products
-108k.s/ mol = B. D f₂ - 2 B.D xe-f
-108 k.s/mol =155 k.s/mol - 2B.Dxe-f
263kJ/mol/2 = B. D xe-f
B.D xef = 131.5 kJ/mol
132 kJ/mol
ΔXe = ΣB.P reactants - Σ B.d products
-108k.s/ mol = B. D f₂ - 2 B.D xe-f
-108 k.s/mol =155 k.s/mol - 2B.Dxe-f
263kJ/mol/2 = B. D xe-f
B.D xef = 131.5 kJ/mol
132 kJ/mol
Answer: Average bond dissociation enthalpy of a (Xe–F) bond is 131.5kJ/mol.
Expatiation:
[tex]Xe(g)+F_2(g)\rightarrow XeF_2(g), \Delta H_{f}=-108kJ/mol[/tex]..(1)
[tex]F_2\rightarrow 2F^-,\Delta H{diss}=155kJ/mol[/tex]..(2)
Subtracting (1) from (2)
[tex]Xe(g)+2F^-(g)\rightarrow XeF_2(g)[/tex]
[tex]\Delta H_{rxn}=-108kJ/mol-155kJ/mol=-263kJ/mol[/tex]
Average bond dissociation enthalpy of a (Xe–F) bond :
[tex]XeF_2\rightarrow Xe+2F^-,\Delat H_{\text{diss. of (Xe-F)}}=263kJ/mol[/tex]
Since there are two (Xe-F) bonds in molecule the average the bond energy of Xe-f bond will be = [tex]\frac{263 kJ/mol}{2}=131.5kJ/mol[/tex]
Hence, Average bond dissociation enthalpy of a (Xe–F) bond is 131.5kJ/mol
