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Given the bond energies N-N , N-H and H-...

Given the bond energies `N-N` , `N-H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

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The correct Answer is:
93
`DeltaH = -2 [3 xx in_(N - H)] +in_(N -= N) + 3 xx in_(H - H) = -2 xx 3 xx 391 + 945 + 436 xx 4 = -93 kJ`
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