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Enthalpy of formation of 2mol of NH(3)(g...

Enthalpy of formation of `2mol` of `NH_(3)(g)` is `-90KJ` , and `DeltaH_(H-H)` and `DeltaH_(N-H)` respectively `435KJ mol^(-1)` and `390KJ mol^(-1)` . The valule of `DeltaH_(N-=N)` is

A

`-472.5KJ`

B

`-945KJmol^(-1)`

C

`472.5KJ`

D

`945KJmol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g),DeltaH=-90KJ`
Now `-90=DeltaH_(N-=N)+3DeltaH_(H-H)+6DeltaH_(N-H)`
`DeltaH_(N-=N)=6DeltaH_(N-H)-3DeltaH_(H-H)-90`
`=6xx390-435-90`
`=945KJ`
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