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Heat of formation of 2 mol of NH(3)(g) i...

Heat of formation of `2 mol` of `NH_(3)(g)` is `= -90 kJ`, bond energies of `H-H` and `N-H` bonds are `435kJ` and `390 kJ mol^(-1)`, respectively. The value of the bond enegry of `N-=N` will be

A

`-472.5kJ`

B

`-945kJ`

C

`472.5kJ`

D

`945kJ" "mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

`N-=+3(H-H)to2underset(H)underset(|)overset(H)overset(|)(N)-H," "DeltaH=-90kJ`
`DeltaH_("reaction")=sum(BE)_("reactants")-sum(BE)_("products")`
`-90=[(BE)_(N-=N)+3(BE)_(H-H)]-[6(BE)_(N-H)]`
`-90=x+3xx435-6xx390`
`x=945" kJ "mol^(-1)`.
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