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

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

$\text{Δ} H_{\text{f}}^{^\circ } = - 4 6 = \displaystyle \sum \text{B.E} - \displaystyle \sum \text{B.E}$

Reactants Product

$\displaystyle \sum \text{B.E} = \displaystyle \sum \text{B.E} + 4 6$

Product Reactants

$= \frac{1}{2} \left(7 1 2\right) + \frac{3}{2} \left(4 3 6\right) + 4 6$

= 356 + 654 + 46 = 1056

Averge B.E for N-H Bond $= \frac{1 0 5 6}{3} = 3 5 2 KJ mol^{- 1}$