Heat of atomisation of `NH_(3)` and `N_(2)H_(4)` are `"x kcal mol"^(-1)` and `"y kcal mol"^(-1)` respectively. Calculate average bond energy of `N-N` bond.

Options

A. `(4y - 3x)/(3) kcal mol^(-1)`
B. `(2y - 3x)/(3) kcal mol^(-1)`
C. `(4y - 3x)/(4) kcal mol^(-1)`
D. `(3y - 4x)/(3) kcal mol^(-1)`

Explanation

To solve the problem of calculating the average bond energy of the N-N bond given the heat of atomization of NH3 and N2H4, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Heat of Atomization**:
- The heat of atomization is the energy required to completely separate a molecule into its individual atoms.
- For NH3 (ammonia), the heat of atomization is given as \( x \) kcal/mol.

2. **Calculate Bond Energy of N-H in NH3**:
- NH3 has three N-H bonds. Therefore, the total bond energy for breaking these bonds is equal to the heat of atomization.
- The relationship can be expressed as:
\[ 3 \times \text{Bond Energy of N-H} = x \]
- Thus, the bond energy of one N-H bond is:
\[ \text{Bond Energy of N-H} = \frac{x}{3} \text{ kcal/mol} \]

3. **Analyze N2H4 (Hydrazine)**:
- N2H4 has four N-H bonds and one N-N bond. The heat of atomization for N2H4 is given as \( y \) kcal/mol.
- The equation for the heat of atomization can be expressed as:
\[ 4 \times \text{Bond Energy of N-H} + \text{Bond Energy of N-N} = y \]
- Substituting the bond energy of N-H from step 2 into this equation gives:
\[ 4 \times \frac{x}{3} + \text{Bond Energy of N-N} = y \]

4. **Rearranging the Equation**:
- Rearranging the equation to solve for the bond energy of the N-N bond:
\[ \text{Bond Energy of N-N} = y - 4 \times \frac{x}{3} \]
- This can be simplified to:
\[ \text{Bond Energy of N-N} = y - \frac{4x}{3} \]

5. **Final Expression**:
- To express it in a single fraction:
\[ \text{Bond Energy of N-N} = \frac{3y - 4x}{3} \text{ kcal/mol} \]

### Conclusion:
The average bond energy of the N-N bond is given by:
\[ \text{Bond Energy of N-N} = \frac{3y - 4x}{3} \text{ kcal/mol} \]