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. `-96 kJ mol^(-1)`
B. `+ 352 kJ mol^(-1)`
C. `+1056 kJ mol^(-1)`
D. `-1102 kJ mol^(-1)`

Answer

B

Explanation

`N_(2)+3H_(2) rarr 2NH_(3)`