1. Write the formation reaction of ammonia

The standard enthalpy of formation of ammonia is for:

$\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$

Given:

$\Delta H_f^\circ(NH_3) = -46.0\ \text{kJ mol}^{-1}$

2. Convert molecules into atoms using given bond dissociation data

Given:

• Enthalpy of formation of $H_2$ from atoms is $-436\ \text{kJ mol}^{-1}$

So, dissociation of $H_2$ is:

$H_2 \rightarrow 2H \qquad \Delta H = +436\ \text{kJ mol}^{-1}$

Hence for $\frac{3}{2}H_2$:

$\frac{3}{2}H_2 \rightarrow 3H \qquad \Delta H = \frac{3}{2}\times 436 = 654\ \text{kJ mol}^{-1}$

Similarly,

• Enthalpy of formation of $N_2$ from atoms is $-712\ \text{kJ mol}^{-1}$

So, dissociation of $N_2$ is:

$N_2 \rightarrow 2N \qquad \Delta H = +712\ \text{kJ mol}^{-1}$

Hence for $\frac{1}{2}N_2$:

$\frac{1}{2}N_2 \rightarrow N \qquad \Delta H = \frac{1}{2}\times 712 = 356\ \text{kJ mol}^{-1}$

3. Total energy required to convert reactants into atoms

$\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightarrow N + 3H$

$\Delta H = 356 + 654 = 1010\ \text{kJ mol}^{-1}$

4. Form ammonia from atoms

$N + 3H \rightarrow NH_3$

Let average bond enthalpy of one $N-H$ bond be $x$.

Since 3 bonds are formed, enthalpy released is:

$-3x$

5. Apply Hess's law

Overall reaction:

$\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightarrow NH_3$

So,

$1010 - 3x = -46$

$-3x = -1056$

$3x = 1056$

$x = 352\ \text{kJ mol}^{-1}$

6. Interpretation

Bond enthalpy is the energy required to break one mole of bonds, so it is taken as positive.

Thus, average bond enthalpy of $N-H$ bond in $NH_3$ is:

$\boxed{+352\ \text{kJ mol}^{-1}}$

7. Check options

• A: $-964$ kJ mol$^{-1}$ ❌
• B: $+352$ kJ mol$^{-1}$ ✅
• C: $+1056$ kJ mol$^{-1}$ ❌ (this is for 3 bonds together)
• D: $-1102$ kJ mol$^{-1}$ ❌