Based on the values of B.E. given, `Delta_(f)H^(@)` of `N_(2)H_(4)` (g) is :
Given : `N-N = 159 " kJ mol"^(-1)," "H-H=436 " kJ mol"^(-1)`
`N-=N = 941" kJ mol"^(-1), " "N-H=398 " kJ mol"^(-1)`

Options

A. 711 kJ `" mol"^(-1)`
B. 62 kJ `" mol"^(-1)`
C. `-98 " kJ mol"^(-1)`
D. `-711 " kJ mol"^(-1)`

Explanation

To find the formation enthalpy (Δ_fH^(@)) of hydrazine (N₂H₄), we will use the bond energies provided and apply Hess's law. The formation reaction for N₂H₄ can be represented as follows:

\[ \text{N}_2(g) + 2\text{H}_2(g) \rightarrow \text{N}_2\text{H}_4(g) \]

### Step 1: Identify the bonds in the reactants and products
- **Reactants:**
- 1 N≡N bond in N₂ (bond energy = 941 kJ/mol)
- 2 H-H bonds in H₂ (bond energy = 436 kJ/mol each)

- **Products:**
- 1 N-N bond in N₂H₄ (bond energy = 159 kJ/mol)
- 4 N-H bonds in N₂H₄ (bond energy = 398 kJ/mol each)

### Step 2: Calculate the total bond energy of the reactants
The total bond energy for the reactants can be calculated as follows:

\[ \text{Total bond energy of reactants} = \text{Bond energy of N≡N} + 2 \times \text{Bond energy of H-H} \]
\[ = 941 \, \text{kJ/mol} + 2 \times 436 \, \text{kJ/mol} \]
\[ = 941 + 872 = 1813 \, \text{kJ/mol} \]

### Step 3: Calculate the total bond energy of the products
The total bond energy for the products can be calculated as follows:

\[ \text{Total bond energy of products} = \text{Bond energy of N-N} + 4 \times \text{Bond energy of N-H} \]
\[ = 159 \, \text{kJ/mol} + 4 \times 398 \, \text{kJ/mol} \]
\[ = 159 + 1592 = 1751 \, \text{kJ/mol} \]

### Step 4: Apply Hess's law to find Δ_fH^(@)
Using Hess's law, the formation enthalpy can be calculated as:

\[ \Delta_fH^(@) = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
\[ = 1813 \, \text{kJ/mol} - 1751 \, \text{kJ/mol} \]
\[ = 62 \, \text{kJ/mol} \]

### Final Answer
The formation enthalpy (Δ_fH^(@)) of N₂H₄ is **62 kJ/mol**.

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