The standard enthalpy of formation of NH3 is -46.0 kJ `mol^(-1)`. If the enthalpy of formation of `H_2` from its atoms is -436 kJ mol and that of `N_2` is -712 kJ mol, the average bond enthalpy of N - H bond in `NH_3` is

Options

A. `-964 kJ mol^(-1)`
B. `+352 kJ mol^(-1)`
C. `+1056 kJ mol^(-1)`
D. `-1102 kJ mol^(-1)`

Answer

B

Explanation

Given , `1/2 N_(2)(g) + 3/2 H_(2)(g) to NH_(3)(g)`
`DeltaH_(f)^(@) = -46.0 kJ mol^(-1)`
`2H(g) to H_(2)(g) DeltaH_(f)^(@) = -436 kJ mol^(-1)`
`2N(g) to N_(2)(g) DeltaH_(f)^(@) = -436 kJ mol^(-1)`
`2N(g) to N_(2)(g) Delta H_(f)^(@) = -712 kJ mol^(-1)`
Assuming X is the bond energy of N - H bond (in `kJ mol^(-1))`
`1/2 xx (712) + 3/2 xx (436) - 3x = -46.0`
`3x = 10556 kJ mol^(-1)`
`3X = 1056 kJ mol^(-1)`
So, `X = 352 kJ mol^(-1)`.