Consider the following data : `Delta_(f)H^(@)(N_(2)H_(4),l)=50kJ//mol,Delta_(f)H^(@)(NH_(3),g)=-46kJ//mol`
`B.E(N-H)=393" kJ//mol and B.E."(H-H)=436kJ//mol`
`Delta_("vap")H(N_(2)H_(4),l)=18kJ//mol`
The N-N bond energy in `N_(2)H_(4)` is :

Options

A. 226 kJ/mol
B. 154 kJ/mol
C. 190 kJ/mol
D. None of these

Explanation

To find the N-N bond energy in hydrazine (N₂H₄), we will use the given data and apply Hess's law. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the formation reaction of NH₃
The formation of ammonia (NH₃) from its elements can be represented as:
\[ \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow 2 \text{NH}_3(g) \]
The enthalpy change for this reaction is given as:
\[ \Delta H_f^\circ(\text{NH}_3) = -46 \text{ kJ/mol} \]

### Step 2: Relate bond energies to the enthalpy change
The bond energies involved in the formation of NH₃ can be expressed as:
- Breaking the N≡N bond in N₂ requires an energy of \( x \) (where \( x \) is the bond energy we want to find).
- Breaking the H-H bonds in H₂ requires \( \frac{3}{2} \times 436 \text{ kJ} \) (since bond energy of H-H is 436 kJ/mol).
- Breaking 3 N-H bonds in NH₃ requires \( 3 \times 393 \text{ kJ} \) (since bond energy of N-H is 393 kJ/mol).

Using Hess's law, we can set up the equation:
\[ -46 = \frac{x}{2} + \frac{3}{2} \times 436 - 3 \times 393 \]

### Step 3: Solve for x
Rearranging the equation gives:
\[ \frac{x}{2} = -46 - \frac{3}{2} \times 436 + 3 \times 393 \]
Now calculate the right side:
1. Calculate \( \frac{3}{2} \times 436 = 654 \text{ kJ} \)
2. Calculate \( 3 \times 393 = 1179 \text{ kJ} \)

Substituting these values:
\[ \frac{x}{2} = -46 - 654 + 1179 \]
\[ \frac{x}{2} = 479 \]
Multiplying both sides by 2:
\[ x = 958 \text{ kJ/mol} \]

### Step 4: Write the formation reaction of N₂H₄
The formation of hydrazine (N₂H₄) from its elements can be represented as:
\[ \text{N}_2(g) + 2 \text{H}_2(g) \rightarrow \text{N}_2H_4(l) \]
The enthalpy change for this reaction is given as:
\[ \Delta H_f^\circ(\text{N}_2H_4) = 50 \text{ kJ/mol} \]

### Step 5: Set up the equation for N₂H₄
Using the bond energies, we can express the enthalpy change for the formation of N₂H₄:
1. Breaking the N≡N bond requires \( x \) (which we found to be 958 kJ).
2. Breaking 4 N-H bonds requires \( 4 \times 393 \text{ kJ} \).
3. Breaking 2 H-H bonds requires \( 2 \times 436 \text{ kJ} \).
4. The vaporization of N₂H₄ requires \( 18 \text{ kJ} \).

Using Hess's law again:
\[ 50 = 958 + 2 \times 436 - 4 \times 393 - 18 - y \]
Where \( y \) is the bond energy of the N-N bond in N₂H₄.

### Step 6: Solve for y
Rearranging the equation gives:
\[ y = 958 + 2 \times 436 - 4 \times 393 - 18 - 50 \]
Calculating the right side:
1. Calculate \( 2 \times 436 = 872 \text{ kJ} \)
2. Calculate \( 4 \times 393 = 1572 \text{ kJ} \)

Substituting these values:
\[ y = 958 + 872 - 1572 - 18 - 50 \]
\[ y = 958 + 872 - 1640 \]
\[ y = 190 \text{ kJ/mol} \]

### Final Answer
The N-N bond energy in N₂H₄ is:
\[ \boxed{190 \text{ kJ/mol}} \]

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