Bond dissociation enthalpies of `H_(2)(g)` and `N_(2)(g)` are `436.0 kJ mol^(-1)` and `941.8 kJ mol^(-1)`, respectively and enthalpy of formation of `NH_(3)(g)` is `-46kJ mol^(-1)`. What is the enthalpy of atomisation of `NH_(3)(g)`?. What is the avergae bond ethalpy of `N-H` bond?

Options

A. `390.3 "kJ mol"^(-1)`
B. `1170.9 "kJ mol"^(-1)`
C. `590 "kJ mol"^(-1)`
D. `720 " kJ mol"^(-1)`

Answer

B

Explanation

`H_(2(g))rarr 2H_((g)) , Delta H=436 KJ mol^(-1)`
`N_(2(g)) rarr 2N_((g)) , Delta H = 941.8 KJ mol^(-1)`
`(1)/(2)N_(2(g))+(3)/(2)H_(2(g))rarr NH_(3(g)) , Delta H =-46 KJ mol^(-1)`
`NH_(3(g))rarr N_((g))+3H_((g)) , Delta H_("atom")= ?`
`eq(iv) =(3)/(2)xx eq(i) +(1)/(2)xx eq(ii)-eq(iii)`
`Delta H_("atom")=(3)/(2)xx(436)+(1)/(2)xx(941.8)-(-46)`
`Delta H_("atom")=1170.9 J mol^(-1)`