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 value of \( \Delta H_{N \equiv N} \) (enthalpy of formation for nitrogen gas), we will use the enthalpy of formation of ammonia and the bond enthalpies provided. Here’s the step-by-step solution:

### Step 1: Write the Reaction for Ammonia Formation
The formation of ammonia (\( NH_3 \)) from nitrogen and hydrogen can be represented by the following balanced chemical equation:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Write the Enthalpy Change for the Reaction
The enthalpy change for the reaction can be expressed as:
\[ \Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants} \]
For this reaction, the enthalpy change can be written as:
\[ \Delta H_{reaction} = 2 \Delta H_{f}(NH_3) - \left( \Delta H_{f}(N_2) + 3 \Delta H_{f}(H_2) \right) \]

### Step 3: Substitute the Known Values
Given:
- \( \Delta H_{f}(NH_3) = -X \)
- \( \Delta H_{f}(H-H) = Y \) (bond enthalpy of H-H)
- \( \Delta H_{f}(N-H) = Z \) (bond enthalpy of N-H)

Substituting these values into the equation:
\[ \Delta H_{reaction} = 2(-X) - \left( \Delta H_{f}(N_2) + 3Y \right) \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \Delta H_{reaction} = -2X - \Delta H_{f}(N_2) - 3Y \]

### Step 5: Relate the Reaction Enthalpy to Bond Enthalpies
The reaction enthalpy can also be expressed in terms of bond enthalpies:
\[ \Delta H_{reaction} = \text{Total bond enthalpy of bonds broken} - \text{Total bond enthalpy of bonds formed} \]
For the bonds broken:
- 1 \( N \equiv N \) bond (which we want to find)
- 3 \( H-H \) bonds

For the bonds formed:
- 6 \( N-H \) bonds (since there are 2 moles of \( NH_3 \) and each has 3 \( N-H \) bonds)

Thus, we can express it as:
\[ \Delta H_{reaction} = \Delta H_{N \equiv N} + 3Y - 6Z \]

### Step 6: Set the Two Expressions Equal
Now we can set the two expressions for \( \Delta H_{reaction} \) equal to each other:
\[ -2X - \Delta H_{f}(N_2) - 3Y = \Delta H_{N \equiv N} + 3Y - 6Z \]

### Step 7: Solve for \( \Delta H_{N \equiv N} \)
Rearranging gives:
\[ \Delta H_{N \equiv N} = -2X - \Delta H_{f}(N_2) - 3Y - 3Y + 6Z \]
\[ \Delta H_{N \equiv N} = -2X - \Delta H_{f}(N_2) - 6Y + 6Z \]

### Step 8: Conclusion
Thus, the value of \( \Delta H_{N \equiv N} \) can be expressed in terms of \( X \), \( Y \), and \( Z \) as:
\[ \Delta H_{N \equiv N} = -2X - \Delta H_{f}(N_2) - 6Y + 6Z \]