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

Options

A. `-872.5 kJ`
B. `-945 kJ`
C. `872.5 kJ`
D. `945 kJ mol^(-1)`

Explanation

`N-= N +3(H-H) rarr 2 overset(H)overset(|)underset(H)underset(|)(N)-H, DeltaH =- 90 kJ`
`DeltaH^(Theta) = sum (BE) (Reactants) -sum (BE) (Products)`
`-90 kJ =[(BE)_(N-=N) +3(BE)_(H_H)] -[6(BE)_(N-H)]`
`-90 kJ = x +3 xx 435 - 6 xx 390`
`:. x = 945 kJ mol^(-1)`