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

The standard enthalpy of formation of NH_(3) is - 46.0 kJ mol^(-1). If

1.	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
Answer» `+ 1056 kJ mol^(-1)` `- 1102 kJ mol^(-1)` `- 964 kJ mol^(-1)` `+ 352 kJ mol^(-1)` Solution :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