Bond energies of `N -= N` H-H, `N - H` are respectively `x_1 , x_2, x` and X_3 `kJ mol^(-1)` . Hence, `DeltaH_(f)^(@)(NH_3)` is:

Options

A. `x_1 + 3x_2 - x_3`
B. `x_1 + x_2 - x_3`
C. `(x_1)/(2) + 3/2 x_(2) - 3x_(3)`
D. `x_1 + 3x_(2) - 6x_(3)`

Explanation

To find the standard enthalpy of formation (ΔH_f^(@)(NH3)) of ammonia (NH3) using the bond energies of the involved molecules, we can follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation for the formation of NH3:**
The formation of ammonia from nitrogen and hydrogen can be represented as:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
To simplify, we can divide the entire equation by 2:
\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g) \]

2. **Identify the bonds broken and formed:**
- Bonds broken:
- 1 N≡N bond (from N2)
- 3 H-H bonds (from H2)
- Bonds formed:
- 3 N-H bonds (in NH3)

3. **Calculate the total bond energy required to break the bonds:**
The total energy required to break the bonds can be expressed as:
\[ \text{Energy required} = \frac{1}{2} \cdot x_1 + \frac{3}{2} \cdot x_2 \]
where \(x_1\) is the bond energy of N≡N, and \(x_2\) is the bond energy of H-H.

4. **Calculate the total bond energy released when forming the bonds:**
The energy released when forming the bonds can be expressed as:
\[ \text{Energy released} = 3 \cdot x_3 \]
where \(x_3\) is the bond energy of N-H.

5. **Apply Hess's law to find ΔH_f:**
According to Hess's law, the standard enthalpy of formation (ΔH_f) can be calculated as:
\[ ΔH_f^(@)(NH_3) = \text{Energy required to break bonds} - \text{Energy released when forming bonds} \]
Substituting the values we calculated:
\[ ΔH_f^(@)(NH_3) = \left(\frac{1}{2} x_1 + \frac{3}{2} x_2\right) - 3 x_3 \]

6. **Final expression for ΔH_f:**
Thus, the final expression for the standard enthalpy of formation of ammonia is:
\[ ΔH_f^(@)(NH_3) = \frac{1}{2} x_1 + \frac{3}{2} x_2 - 3 x_3 \]