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 follow these steps:

### Step 1: Write the balanced chemical equation for the formation of ammonia.
The formation of ammonia from nitrogen and hydrogen can be represented as:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \]

### Step 2: Identify the bond enthalpies given in the problem.
- Bond enthalpy of \( H_2(g) \) = 436 kJ/mol
- Bond enthalpy of \( N_2(g) \) = 941.3 kJ/mol
- Heat of reaction for the formation of ammonia, \( \Delta H^\circ_{NH_3} \) = -46 kJ/mol

### Step 3: Calculate the total bond enthalpy of the reactants.
The total bond enthalpy for the reactants can be calculated as follows:
- For \( \frac{1}{2} N_2 \):
\[ \text{Bond enthalpy} = \frac{1}{2} \times 941.3 \text{ kJ/mol} = 470.65 \text{ kJ/mol} \]
- For \( \frac{3}{2} H_2 \):
\[ \text{Bond enthalpy} = \frac{3}{2} \times 436 \text{ kJ/mol} = 654 \text{ kJ/mol} \]

Adding these together gives:
\[ \text{Total bond enthalpy of reactants} = 470.65 + 654 = 1124.65 \text{ kJ/mol} \]

### Step 4: Use the heat of reaction to find the total bond enthalpy of the products.
The heat of reaction is given by the equation:
\[ \Delta H = \text{Bond enthalpy of reactants} - \text{Bond enthalpy of products} \]

Rearranging gives us:
\[ \text{Bond enthalpy of products} = \text{Bond enthalpy of reactants} - \Delta H \]

Substituting the values:
\[ \text{Bond enthalpy of products} = 1124.65 \text{ kJ/mol} - (-46 \text{ kJ/mol}) \]
\[ \text{Bond enthalpy of products} = 1124.65 + 46 = 1170.65 \text{ kJ/mol} \]

### Step 5: Calculate the average bond enthalpy of an N-H bond.
Since there are 3 N-H bonds in ammonia, the average bond enthalpy of one N-H bond is:
\[ \text{Average bond enthalpy of N-H} = \frac{\text{Bond enthalpy of products}}{3} \]
\[ \text{Average bond enthalpy of N-H} = \frac{1170.65 \text{ kJ/mol}}{3} = 390.21 \text{ kJ/mol} \]

### Final Answer:
The average bond enthalpy of an N-H bond in ammonia is approximately **390.21 kJ/mol**.

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