The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` is

Options

A. `-1102` kJ/mol
B. `-964`kJ/mol
C. `+352` kJ/mol
D. `+1056`kJ/mol

Answer

B

Explanation

Given `1/2N_(2)+3/2H_(2)hArrNH_(3),`
`DeltaH_(f)=-46.0` kJ/mol
`H+HhArrH_(2),DeltaH_(f)=-436` kJ/mol
`N+NhArrN_(2),DeltaH_(f)=-712` kJ/mol
`DeltaH_(f)(NH_(3)=)1/2DeltaH_(N-N)+3/2DeltaH_(H-H)-DeltaH_(N-F)`
`-46=1/2(-712)3/2(-436)DeltaH_(N-F)`
On calculation
`DeltaH_(N-F)=-964` kJ/mol