Use bond energies to calculate the enthalpy of formation of nh3 in kj/mol.

Solution

ΔH=[ΔH
(N≡N)
​
+3×ΔH
(H−H)
​
]−[6ΔH
(N−H)
​
]
945+3×435.0−6×389.0=−83.64kJ
Heat formation of NH
3
​
=
2
ΔH
​
=−
2
83.64
​
=−41.82kJmol
−1