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jolli1 [7]
3 years ago
15

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
Chemistry
2 answers:
S_A_V [24]3 years ago
4 0
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
lukranit [14]3 years ago
4 0

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

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