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. `-9 6 4 K J m o l^(-1)`
B. `+ 3 5 2 K J m o l^(-1)`
C. `+ 1 0 5 6 K J m o l ^(-1)`
D. `-1 1 0 2 K J m o l ^(-1)`

Answer

B

Explanation

`1/2 N_2(g) + 3/2 H_2(g) to NH_(3)(g), Delta_(f )H = - 46 kJ mol^(-1)`
Bond enthalpy of `H_2 = 436 kJ mol^(-1)`
[+ve sign is taken because energy is supplied to break the H–H bond into its atoms]
Similarly, bond enthalpy of `N_2 = _712 kJ mol^(-1)`
`Delta_(f)H = [1/2 BE(N_2) + 3/2 BE(H_2)] - BE (N - N)`
`-46 = [1/2 xx 712 + 3/2 xx 436] - 3BE(N - H)`
`-46 = (356 + 654) - 3BE(N - H)`
`3BE (N - H) = (1010) + 46`
`3BE (N - H) = 1056`
`Be(N - H) = 1056//3 = 352 kJ mol^(-1)`.