The standard enthalpy of formation of `NH_(3)` is `-46 kJ mol^(-1).` If the enthalpy of formation of H_(2), from its atoms is `-436 kJ mol^(-1)` and that of `N_(2) is - 712 kJ mol^(-1).` the average bond enthalpy of N-H bond in `NH_(3)` is:

Options

A. `+ 1066 kJ mol ^(-1)`
B. `- 11 02 kJ mol ^(-1)`
C. `- 964 kJ mol ^(-1)`
D. `+ 352 kJ mol ^(-1)`

Explanation

`(1)/(2) ( N -= N) + (3)/(2) ( H - H) to NH _(3) Delta H = - 46kJ mol ^(-1)`
`Delta H _(f) ^(0) (N -= N) =- 712 mol ^(-1) Delta H _(f) ^(0) (H - H) =- 436 kJ mol ^(-1)`
Let `Delta H _(f)^(0) of (N - H) = x kJ mol ^(-1) Delta H ^(0) = sum Delta H _("f products") ^(0) - sum Delta H _("f Reactants") ^(@)`
`sum BE _("reactants") - sum BE _("products") = BE _( N - H) = Delta H = 352 kJ mol ^(-1)` (Bond dissociation enthalpy is positive)