The standard enthalpy of formation of $$\text{NH}_3$$ is $$-46.0$$ kJ mol$$^{-1}$$. If the enthalpy of formation of $$\text{H}_2$$ from its atoms is $$-436$$ kJ mol$$^{-1}$$ and that of $$\text{N}_2$$ is $$-712$$ kJ mol$$^{-1}$$, the average bond enthalpy of N$$-$$H bond in $$\text{NH}_3$$ is

• Standard Enthalpy of Formation of $$\text{NH}_3$$:

  This corresponds to synthesizing $$1 \text{ mole}$$ of ammonia gas from its constituent elements in their standard states:

  $$\frac{1}{2}\text{N}_{2(g)} + \frac{3}{2}\text{H}_{2(g)} \rightarrow \text{NH}_{3(g)} \quad \Delta H_f^\circ = -46.0 \text{ kJ mol}^{-1}$$




• Enthalpy of Formation of $$\text{H}_2$$ from its Atoms (Bond Formation):

  The problem gives the enthalpy of forming $$\text{H}_2$$ from gaseous atoms as $$-436 \text{ kJ mol}^{-1}$$. Since bond dissociation is the exact reverse process, the bond dissociation enthalpy of $$\text{H--H}$$ ($$\Delta H_{\text{diss}}^\circ$$) is positive:

  $$\text{H}_{2(g)} \rightarrow 2\text{H}_{(g)} \quad \Delta H^\circ = +436 \text{ kJ mol}^{-1}$$




• Enthalpy of Formation of $$\text{N}_2$$ from its Atoms (Bond Formation):

  Similarly, the enthalpy of forming $$\text{N}_2$$ from individual nitrogen atoms is given as $$-712 \text{ kJ mol}^{-1}$$. Therefore, the bond dissociation enthalpy of the $$\text{N}\equiv\text{N}$$ triple bond is:

  $$\text{N}_{2(g)} \rightarrow 2\text{N}_{(g)} \quad \Delta H^\circ = +712 \text{ kJ mol}^{-1}$$

Using the principle that the enthalpy of a reaction can be calculated by subtracting the total bond enthalpies of the products from the total bond enthalpies of the reactants:

$$\Delta H_f^\circ = \sum \Delta H_{\text{bond}} \text{(Reactants)} - \sum \Delta H_{\text{bond}} \text{(Products)}$$

For the formation reaction of $$\text{NH}_3$$:

$$\Delta H_f^\circ = \left[ \frac{1}{2} \Delta H_{\text{bond}}(\text{N}\equiv\text{N}) + \frac{3}{2} \Delta H_{\text{bond}}(\text{H--H}) \right] - \left[ 3 \times \Delta H_{\text{bond}}(\text{N--H}) \right]$$

• Substitute the known values into the equation:

  $$-46.0 = \left[ \frac{1}{2}(712) + \frac{3}{2}(436) \right] - 3 \times \Delta H_{\text{bond}}(\text{N--H})$$




• Simplify the reactant bond enthalpies:

  $$\frac{1}{2}(712) = 356$$

  $$\frac{3}{2}(436) = 3 \times 218 = 654$$

  $$\text{Total Reactant Bond Energy} = 356 + 654 = 1010 \text{ kJ}$$




• Solve for the total product bond enthalpy term:

  $$-46.0 = 1010 - 3 \times \Delta H_{\text{bond}}(\text{N--H})$$

  $$3 \times \Delta H_{\text{bond}}(\text{N--H}) = 1010 + 46.0$$

  $$3 \times \Delta H_{\text{bond}}(\text{N--H}) = 1056 \text{ kJ}$$




• Isolate the average single $$\text{N--H}$$ bond enthalpy:

  $$\Delta H_{\text{bond}}(\text{N--H}) = \frac{1056}{3} = +352 \text{ kJ mol}^{-1}$$

Conclusion:

The average bond enthalpy required to break one mole of $$\text{N--H}$$ single bonds in ammonia is calculated to be $$+352 \text{ kJ mol}^{-1}$$.

Answer: Option B — $$+352 \text{ kJ mol}^{-1}$$