Enthalpy of formation of `NH_(3)` is -X kJ and `Delta H_(H-H)`, `Delta H_(N-H)` are respectively Y kJ`mol^(-1)` and Z kj`mol^(-1)`. The value of `Delta H_(N = N) is`

Options

A. `Y -6Z + X/3`
B. `-3Y + 6Z +2X`
C. `3Y + 6Z + X`
D. `Y + 6X + Z`

Answer

To find the bond energy of the nitrogen double bond (N≡N), we can analyze the reaction for the formation of ammonia (NH₃) and apply Hess's law.

### Step-by-Step Solution:

1. **Write the Reaction for Formation of Ammonia:**
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Identify the Enthalpy of Formation:**
The enthalpy of formation of ammonia (NH₃) is given as -X kJ. Therefore, for 2 moles of NH₃, the total enthalpy change will be:
\[ \Delta H = 2 \times (-X) = -2X \text{ kJ} \]

3. **Calculate the Bond Energies:**
- The bond energy of the nitrogen-nitrogen triple bond (N≡N) is what we need to find.
- The bond energy of hydrogen (H-H) is given as Y kJ/mol. Since there are 3 moles of H₂, the total energy for breaking 3 moles of H-H bonds is:
\[ 3Y \text{ kJ} \]
- The bond energy of nitrogen-hydrogen (N-H) bonds is given as Z kJ/mol. In 2 moles of NH₃, there are 6 N-H bonds, so the total energy for forming 2 moles of NH₃ is:
\[ 6Z \text{ kJ} \]

4. **Apply Hess's Law:**
According to Hess's law, the enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{Bond energy of reactants} - \text{Bond energy of products} \]
Substituting the values we have:
\[ -2X = \text{Bond energy of } N_2 + 3Y - 6Z \]
Rearranging gives:
\[ \text{Bond energy of } N_2 = -2X + 6Z - 3Y \]

5. **Final Expression:**
The bond energy of the nitrogen double bond (N≡N) can be expressed as:
\[ \Delta H_{N≡N} = -2X + 6Z - 3Y \]

### Conclusion:
The value of \(\Delta H_{N≡N}\) is given by the expression:
\[ \Delta H_{N≡N} = -2X + 6Z - 3Y \]