Chemistry - Thermochemistry Question with Solution | TestHub

For The Reaction

$\frac{1}{2}{\text{N}}_2\left(\text{g}\right)+\frac{3}{2}{\text{H}}_2\left(\text{g}\right)\to {\text{NH}}_3\left(\text{g}\right);\text{Δ}{\text{H}}_{\text{f}}^{\circ }=-46{KJ\; mol}^{-1}$

$\text{Δ}{H}_{\text{f}}^{\circ }=-46=\sum \text{B.E}-\sum \text{B.E}$
Reactants      Product

$\sum \text{B.E}=\sum \text{B.E}+46$
Product        Reactants

$=\frac{1}{2}\left(712\right)+\frac{3}{2}\left(436\right)+46$

= 356 + 654 + 46 = 1056

Averge B.E for N-H Bond$=\frac{1056}{3}=352{KJ\; mol}^{-1}$