Question

The enthalpy of formation of xef2(g) is –108 kj mol–1 and the bond dissociation enthalpy of the f–f bond is 155 kj mol–1 . what is the average bond dissociation enthalpy of a xe–f bond

Answer 1

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

Answer 2

Answer: Average bond dissociation enthalpy of a (Xe–F) bond is 131.5kJ/mol.

Expatiation:

$Xe(g)+F_2(g)\rightarrow XeF_2(g), \Delta H_{f}=-108kJ/mol$..(1)

$F_2\rightarrow 2F^-,\Delta H{diss}=155kJ/mol$..(2)

Subtracting (1) from (2)

$Xe(g)+2F^-(g)\rightarrow XeF_2(g)$

$\Delta H_{rxn}=-108kJ/mol-155kJ/mol=-263kJ/mol$

Average bond dissociation enthalpy of a (Xe–F) bond :

$XeF_2\rightarrow Xe+2F^-,\Delat H_{\text{diss. of (Xe-F)}}=263kJ/mol$

Since there are two (Xe-F) bonds in molecule the average the bond energy of Xe-f bond will be = $\frac{263 kJ/mol}{2}=131.5kJ/mol$

Hence, Average bond dissociation enthalpy of a (Xe–F) bond is 131.5kJ/mol