`NH_(3)(g) + 3Cl_(2)(g) rarr NCl_(3)(g) + 3HCl(g), " "DeltaH_(1)`
`N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g), " "Delta H_(2)`
`H_(2)(g) + Cl_(2)(g) rarr 2HCl(g), " "Delta H_(3)`
The heat of formation of NCl3(g) in the terms of `DeltaH_(1), DeltaH_2 and DeltaH_(3)` is :

Options

A. `DeltaH_(f) = -DeltaH_(1) + (DeltaH_2)/(2) - 3/2 DeltaH_(3)`
B. `DeltaH_(f) = DeltaH_(1) + (DeltaH_2)/(2) - 3/2 DeltaH_(3)`
C. `DeltaH_(f) = DeltaH_(1) - (DeltaH_2)/(2) - 3/2 DeltaH_(3)`
D. None of these

Explanation

To find the heat of formation of NCl₃(g) in terms of ΔH₁, ΔH₂, and ΔH₃, we will analyze the given reactions and rearrange them accordingly.

### Step-by-Step Solution:

1. **Identify the Reactions**:
We have the following reactions:
- Reaction 1: \( \text{NH}_3(g) + 3\text{Cl}_2(g) \rightarrow \text{NCl}_3(g) + 3\text{HCl}(g) \) with ΔH = ΔH₁
- Reaction 2: \( \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \) with ΔH = ΔH₂
- Reaction 3: \( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \) with ΔH = ΔH₃

2. **Understand Heat of Formation**:
The heat of formation of a compound is the change in enthalpy when one mole of that compound is formed from its elements in their standard states. For NCl₃, the standard states are:
- Nitrogen: \( \text{N}_2(g) \)
- Chlorine: \( \text{Cl}_2(g) \)

Therefore, the reaction for the formation of NCl₃ from its elements is:
\[ \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{Cl}_2(g) \rightarrow \text{NCl}_3(g) \]

3. **Rearranging the Reactions**:
- **From Reaction 1**: We can use it directly as it produces NCl₃.
- **From Reaction 2**: We need to form 1 mole of NH₃. So, we divide the entire reaction by 2:
\[ \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \text{(ΔH = } \frac{1}{2} \Delta H₂\text{)} \]
- **From Reaction 3**: We need to reverse this reaction and adjust it to produce 3 moles of HCl. The original reaction produces 2 moles of HCl, so we reverse it and multiply by \( \frac{3}{2} \):
\[ 3\text{HCl}(g) \rightarrow \frac{3}{2} \text{H}_2(g) + \frac{3}{2} \text{Cl}_2(g) \quad \text{(ΔH = } -\frac{3}{2} \Delta H₃\text{)} \]

4. **Combining the Reactions**:
Now we can combine these reactions:
- From Reaction 1: \( \text{NH}_3(g) + 3\text{Cl}_2(g) \rightarrow \text{NCl}_3(g) + 3\text{HCl}(g) \) (ΔH = ΔH₁)
- From modified Reaction 2: \( \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g) \) (ΔH = \( \frac{1}{2} \Delta H₂ \))
- From modified Reaction 3: \( 3\text{HCl}(g) \rightarrow \frac{3}{2} \text{H}_2(g) + \frac{3}{2} \text{Cl}_2(g) \) (ΔH = \( -\frac{3}{2} \Delta H₃ \))

When we add these reactions together, the HCl cancels out, and we are left with:
\[ \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{Cl}_2(g) \rightarrow \text{NCl}_3(g) \]

5. **Calculating ΔH for the Formation of NCl₃**:
The overall change in enthalpy for the formation of NCl₃ is:
\[ \Delta H_{\text{formation of NCl}_3} = \Delta H₁ + \frac{1}{2} \Delta H₂ - \frac{3}{2} \Delta H₃ \]

### Final Answer:
The heat of formation of NCl₃(g) in terms of ΔH₁, ΔH₂, and ΔH₃ is:
\[ \Delta H_{\text{formation of NCl}_3} = \Delta H₁ + \frac{1}{2} \Delta H₂ - \frac{3}{2} \Delta H₃ \]