the bond enthalpy of `H_(2)(g)` is 436 ` kJ "mol"^(-1)` and that of `N_(2)(g)` is 941.3 `kJ"mol"^(-1)`. Calculate the average bond enthalpy of ann N-H bond in ammonia, `Delta_(t)^(@)(NH_(3)) =-46 kJ "mol"^(-1)`

Answer

To calculate the average bond enthalpy of an N-H bond in ammonia (NH₃), we will use the given bond enthalpies of H₂ and N₂, along with the enthalpy change for the formation of ammonia.

### Step-by-Step Solution:

1. **Write the Formation Reaction of Ammonia:**
The formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) can be represented as:
\[ \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g) \]

2. **Identify Given Values:**
- Bond enthalpy of H₂ = 436 kJ/mol
- Bond enthalpy of N₂ = 941.3 kJ/mol
- Enthalpy change for the reaction, ΔH = -46 kJ/mol

3. **Use the Bond Enthalpy Formula:**
The bond enthalpy can be calculated using the formula:
\[ \Delta H = \text{Bond enthalpy of reactants} - \text{Bond enthalpy of products} \]

4. **Calculate the Bond Enthalpy of Reactants:**
For the reactants:
- For N₂: Since we have \(\frac{1}{2}\) mole of N₂, the bond enthalpy contribution is:
\[ \frac{1}{2} \times 941.3 \, \text{kJ/mol} = 470.65 \, \text{kJ} \]
- For H₂: Since we have \(\frac{3}{2}\) moles of H₂, the bond enthalpy contribution is:
\[ \frac{3}{2} \times 436 \, \text{kJ/mol} = 654 \, \text{kJ} \]
- Total bond enthalpy of reactants:
\[ 470.65 + 654 = 1124.65 \, \text{kJ} \]

5. **Set Up the Equation:**
Now, substituting into the bond enthalpy formula:
\[ -46 = 1124.65 - 3 \times \text{Bond enthalpy of N-H} \]

6. **Rearrange to Solve for N-H Bond Enthalpy:**
Rearranging the equation gives:
\[ 3 \times \text{Bond enthalpy of N-H} = 1124.65 + 46 \]
\[ 3 \times \text{Bond enthalpy of N-H} = 1170.65 \]
\[ \text{Bond enthalpy of N-H} = \frac{1170.65}{3} = 390.22 \, \text{kJ/mol} \]

### Final Answer:
The average bond enthalpy of an N-H bond in ammonia is approximately **390.22 kJ/mol**.