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?

Explanation

`N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH=-2xx46kJ//mol`
`DeltaH=sum(BE)_(R)-sum(BE)_(P)`
`=(941.8+3xx436)-(6x)=-2xx46` (Here x=BE of N-H bonds)
`x=390kJ" "mol^(-1)`
`NH_(3)toN+3(H)`
Heat of atomization `=3xx390.3=1170.9kJ" "mol^(-1)`.