Given that Bond energy of `H_(2)` and `N_(2)` are `400KJmol^(-1)` and `240KJmol^(-1)` respectively and `DeltaH_(f)` of `NH_(3)` is `-120KJmol^(-1)` calculate the bond energy of `N-H` bond :-

Options

A. `300KJmol^(-1)`
B. `250KJmol^(-1)`
C. `410KJmol^(-1)`
D. `280KJmol^(-1)`

Explanation

To calculate the bond energy of the N-H bond in NH3, we can follow these steps:

### Step 1: Write the balanced chemical equation for the formation of NH3
The formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) can be represented as:
\[ \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightarrow NH_3 \]

### Step 2: Identify the bond energies and enthalpy of formation
We are given:
- Bond energy of \( H_2 = 400 \, \text{kJ/mol} \)
- Bond energy of \( N_2 = 240 \, \text{kJ/mol} \)
- Enthalpy of formation \( \Delta H_f (NH_3) = -120 \, \text{kJ/mol} \)

### Step 3: Write the equation for the enthalpy change in terms of bond energies
The enthalpy change (\( \Delta H \)) for the reaction can be expressed as:
\[ \Delta H = \text{Bond energy of reactants} - \text{Bond energy of products} \]

### Step 4: Express the bond energies in the equation
For the reactants:
- The bond energy of \( N_2 \) is \( \frac{1}{2} \times 240 \, \text{kJ/mol} \)
- The bond energy of \( H_2 \) is \( \frac{3}{2} \times 400 \, \text{kJ/mol} \)

For the products:
- The bond energy of \( NH_3 \) is \( 3 \times \text{Bond energy of N-H} \)

### Step 5: Set up the equation using the known values
Substituting the values into the equation:
\[ -120 = \left( \frac{1}{2} \times 240 + \frac{3}{2} \times 400 \right) - 3 \times \text{Bond energy of N-H} \]

### Step 6: Simplify the equation
Calculating the bond energies:
\[ \frac{1}{2} \times 240 = 120 \, \text{kJ/mol} \]
\[ \frac{3}{2} \times 400 = 600 \, \text{kJ/mol} \]
Thus, the equation becomes:
\[ -120 = (120 + 600) - 3 \times \text{Bond energy of N-H} \]
\[ -120 = 720 - 3 \times \text{Bond energy of N-H} \]

### Step 7: Rearranging the equation to solve for the bond energy of N-H
\[ -120 - 720 = -3 \times \text{Bond energy of N-H} \]
\[ -840 = -3 \times \text{Bond energy of N-H} \]
\[ \text{Bond energy of N-H} = \frac{840}{3} = 280 \, \text{kJ/mol} \]

### Final Answer:
The bond energy of the N-H bond is \( 280 \, \text{kJ/mol} \).

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