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The standard enthalpy of formation of NH...

The standard enthalpy of formation of `NH_(3)` is `-46.0KJ mol^(-1)` . If the enthalpy of formation of `H_(2)` from its atoms is `-436KJ mol^(-1)` and that of `N_(2)` is `-712KJ mol^(-1)` , the average bond enthalpy of `N-H` bond in `NH_(3)` is

Text Solution

Verified by Experts

The correct Answer is:
`-964`
Given `1/2 N_(2) + 3/2 H_(2) `DeltaH_(f) =-46.0 kJ`/mol
`H + H `N + N_(2) `DeltaH_(f) (NH_(3)) = 1/2 DeltaH_(N-N) + 3/2 DeltaH_(H-H) - DeltaH_(N-H)`
`=-46 = 1/2 (-712) + 3/2(-436) -DeltaH_(N-H)`
On calculation
`DeltaH_(N-H) = -964` kJ/mol
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