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 of ammonia (NH₃), we will use the bond energies provided for the bonds involved in the reaction. Here’s the step-by-step solution:

### Step 1: Write the Formation Reaction
The formation of ammonia from its elements in their standard states can be represented as:
\[ \frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \]

### Step 2: Identify the Bonds Broken and Formed
In this reaction:
- **Bonds Broken**:
- 1 bond of \( N \equiv N \) (triple bond) from \( \frac{1}{2} N_2 \)
- 3 bonds of \( H-H \) (single bonds) from \( \frac{3}{2} H_2 \)

- **Bonds Formed**:
- 3 bonds of \( N-H \) (single bonds) in \( NH_3 \)

### Step 3: Write the Expression for ΔH
The standard enthalpy change for the reaction (ΔH) can be expressed in terms of the bond energies:
\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

### Step 4: Substitute the Bond Energies
Using the bond energies provided:
- Bond energy of \( N \equiv N \) = \( x_1 \)
- Bond energy of \( H-H \) = \( x_2 \)
- Bond energy of \( N-H \) = \( x_3 \)

The expression for ΔH becomes:
\[ \Delta H = \left( \frac{1}{2} x_1 + \frac{3}{2} x_2 \right) - (3 x_3) \]

### Step 5: Simplify the Expression
Combining the terms gives:
\[ \Delta H = \frac{1}{2} x_1 + \frac{3}{2} x_2 - 3 x_3 \]

### Final Result
Thus, the standard enthalpy of formation of ammonia (ΔH_f^°(NH₃)) is:
\[ \Delta H_f^°(NH_3) = \frac{1}{2} x_1 + \frac{3}{2} x_2 - 3 x_3 \]

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