`DeltaH_(f)^(@)` of `NF_(3)` is -113 kJ `mol^(-1)` and N-F bond energy is 273.0 kJ `mol^(-1)`. If `N-=N` and F-F bond energies are in the rates 6:1, their magnetudes will be

Options

A. 780.0 kJ `mol^(-1)`, 130 kJ `mol^(-1)`
B. 840 kJ `mol^(-1)`, 140 kJ `mol^(-1)`
C. 950.0 kJ `mol^(-1)`, 158.3 kJ `mol^(-1)`
D. 941.3 kJ `mol^(-1)`, 156 kJ `mol^(-1)`

Answer

D

Explanation

(d) `(1)/(2)N_(2)(g)+(3)/(2)F_(2)(g) to NF_(3),DeltaH=-113kJ`
or `DeltaH_(N-=N)+(3)/(2)DeltaH_(F-F)-3DeltaH_(N-F)=-113kJ`
Let x kJ `mol^(-1)` be the bond energy of F-F bond then bond energy of `N-=N` bond =6x
`therefore (1)/(2)xx6x+(3)/(2)xx x-3xx273=-113kJ`
On solving,
`x=156.9kJ" "mol^(-1) and N-=N` bond energy`=6xx156.9=941.4kJ" "mol^(-1)`