The enthalpy change `(DeltaH)` for the process, `N_(2)H_(4)(g)to 2N(g)+4H(g)` is
is 1724 kJ `mol^(-1)`. If the bond energy of N-H bond in ammonia is 391 kJ `mol^(-1)`, what is the bond energy for N-N bond in `N_(2)H_(4)`?

Options

A. `391 kJ mol^(-1)`
B. `160 kJ mol^(-1)`
C. `1173 kJ mol^(-1)`
D. `320 kJ mol^(-1)`

Explanation

To find the bond energy for the N-N bond in hydrazine (N₂H₄), we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ N_2H_4(g) \rightarrow 2N(g) + 4H(g) \]
This indicates that one mole of hydrazine decomposes into two moles of nitrogen gas and four moles of hydrogen gas.

### Step 2: Identify Bonds in N₂H₄
In the molecule N₂H₄, there is:
- 1 N-N bond
- 4 N-H bonds

### Step 3: Write the Enthalpy Change Equation
The enthalpy change (ΔH) for the reaction can be expressed in terms of the bond energies:
\[ \Delta H = \text{Bond energy of N-N bond} + 4 \times \text{Bond energy of N-H bond} \]
Given:
- ΔH = 1724 kJ/mol
- Bond energy of N-H bond = 391 kJ/mol

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 1724 = \text{Bond energy of N-N bond} + 4 \times 391 \]

### Step 5: Calculate the Total Energy for N-H Bonds
Calculate the total energy for breaking the 4 N-H bonds:
\[ 4 \times 391 = 1564 \text{ kJ} \]

### Step 6: Solve for the Bond Energy of N-N Bond
Now substitute this value back into the equation:
\[ 1724 = \text{Bond energy of N-N bond} + 1564 \]
Rearranging gives:
\[ \text{Bond energy of N-N bond} = 1724 - 1564 \]
\[ \text{Bond energy of N-N bond} = 160 \text{ kJ/mol} \]

### Conclusion
The bond energy for the N-N bond in N₂H₄ is **160 kJ/mol**.

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