Question

The standard enthalpy of formation of NH₃ is - 46.0 kJ/mol. If bond enthalpy of H₂ is - 436 kJ/mol and that of N₂ is - 712 kJ/mol , the average bond enthalpy of N - H bond in NH₃ is :

• A. - 964 kJ/mol
• B. - 1102 kJ/mol
• C. + 1056 kJ/mol
• D. + 352 kJ/mol

Answer 1

Answer: (D)

For The Reaction

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

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

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 $=\frac{1056}{3}=352KJmo{l}^{-1}$