The standard enthalpy of formation of `NH_(3)` is `- 46.0 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 is `NH_(3)` is

Options

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

Answer

D

Explanation

The required equation is
(i) `NH_(3) (g) rarr N (g) + 3H (g)`
The given equation are
(ii) `(1)/(2)N_(2) (g) + (3)/(2) H_(2) (g) rarr NH_(3) (g) Delta H = - 46 kJ`
(iii) `2H (g) rarr H_(2) (g) Delta H = - 436 kJ`
(iv) `2N (g) rarr N_(2) (g) Delta H = - 712 kJ`
Multiplying eq. (iv) by `(1)/(2)` and eq. (iii) by `(3)/(2)` and add
(v) `N (g) + 3H (g) rarr (1)/(2) N_(2) (g) + 3H_(2) (g) Delta H = - 1010 kJ`
Add eq. (iv) and (ii)
`N (g) + 3H (g) rarr NH_(3) (g) Delta H = - 1056 kJ`
or `N (g) + 3H (g) rarr NH_(3) (g) Delta H = 1056 kJ`
This gives enthalpy of dissociation of 3 N - H bonds
:. Average bond enthalpy of N - H = `(1056)/(3)`
= 352 kJ