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 enthalpy of formation (ΔfH°) of hydrazine (N2H4) gas, we will use the bond energy values provided. The reaction for the formation of N2H4 from nitrogen and hydrogen 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:**
- N2 has one N≡N (triple bond).
- H2 has two H-H (single bonds).

- **Products:**
- N2H4 has one N-N (single bond) and four N-H (single bonds).

### Step 2: Write the bond energy equation
The enthalpy of formation can be calculated using the formula:

\[ \Delta_f H^\circ = \text{(Bond Energy of Reactants)} - \text{(Bond Energy of Products)} \]

### Step 3: Substitute the bond energies
Using the provided bond energies:
- Bond energy of N≡N = 941 kJ/mol
- Bond energy of H-H = 436 kJ/mol
- Bond energy of N-N = 159 kJ/mol
- Bond energy of N-H = 398 kJ/mol

Now we can substitute these values into the equation:

\[ \Delta_f H^\circ = \left( 941 + 2 \times 436 \right) - \left( 159 + 4 \times 398 \right) \]

### Step 4: Calculate the bond energy of reactants
Calculating the bond energy of the reactants:

\[ \text{Bond Energy of Reactants} = 941 + 2 \times 436 = 941 + 872 = 1813 \text{ kJ/mol} \]

### Step 5: Calculate the bond energy of products
Calculating the bond energy of the products:

\[ \text{Bond Energy of Products} = 159 + 4 \times 398 = 159 + 1592 = 1751 \text{ kJ/mol} \]

### Step 6: Calculate ΔfH°
Now, substituting back into the equation:

\[ \Delta_f H^\circ = 1813 - 1751 = 62 \text{ kJ/mol} \]

### Final Answer
Thus, the enthalpy of formation of N2H4 (g) is:

\[ \Delta_f H^\circ = 62 \text{ kJ/mol} \]

---