1.

The standard enthalpy of formation of NH_(3) is -46.0kJmol^(-1). If the enthalpy of formation of H_(2) from its atoms is -436kJmol^(-1), the average bond enthalpy of N-H bond in NH_(3) is

Answer»

`-964kJmol^(-1)`
`-352kJmol^(-1)`
`+1056kJmol^(-1)`
`-1102kJmol^(-1)`

Solution :Given, `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) to NH_(3)(g),DeltaH_(f)^(@)=-46.0" kJ "MOL^(-1)`
`2H(g)toH_(2)(g),DeltaH_(f)^(@)=-436" kJ "Mol^(-1)`
`2N(g)toN_(2)(g),DeltaH_(f)^(@)=-712" kJ "mol^(-1)`
Bond enthalpy of `NH_(3)`
`=3BE_(N-H)[(1)/(2)BE_(N-N)+(3)/(2)BE_(H-H)]`
`-46=3N-N-[-(1)/(2)xx712-(3)/(2)xx436]`
`3N-H=-1056" kJ "mol^(-1)`
`N-H=(1056)/(3)=-352" kJ "mol^(-1)`


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