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^(-1)`
B. `-964 kJ mol^(-1)`
C. `-352 kJ mol^(-1)`
D. `+1056 kJ mol^(-1)`

Answer

C

Explanation

`(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)toNH_(3)(g)`
`DeltaH_(f)^(@)=-46.0 kJ mol^(-1)`
`2H(g)toH_(2)(g)`, `DeltaH_(f)^(@)=-436 kJ mol^(-1)`
`2N(g)toN_(2)(g)` , `DeltaH_(f)^(@)=-712 kJ mol^(-1)`
Assuming `x` as the bond energy of `N-H` bond (in `kJ mol^(-1)`)
`(1)/(2)xx(-712)+(3)/(2)xx(-436)-3x=-46.0`
`:. 3x=1056 kJ mol^(-1)`
`x=352 kJ mol^(-1)`