Chemical BondingEasy
Question
What is bond enthalpy of XeF bond ?
XeF4(g) → Xe+(g) +F-(g) + F2(g) + F(g) ;
ᐃrH = 292 KCal/mol
Given :
I.E. of Xe = 279 KCal/mol.
B.E. (F-F) = 38 KCal/mol.
E.A of F = 85 KCal/mol.
XeF4(g) → Xe+(g) +F-(g) + F2(g) + F(g) ;
ᐃrH = 292 KCal/mol
Given :
I.E. of Xe = 279 KCal/mol.
B.E. (F-F) = 38 KCal/mol.
E.A of F = 85 KCal/mol.
Options
A.24 KCal/mol
B.34 KCal/mol
C.8.5 KCal/mol
D.54 KCal/mol
Solution
ᐃrH = Absorbed - Released
⇒ 292 = (4x + 279) - (38 + 85)
⇒ x = 34 KCal/mol.
⇒ 292 = (4x + 279) - (38 + 85)
⇒ x = 34 KCal/mol.
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