`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 step by step.

### Step 1: Write down the reactions and their respective ΔH values.

1. **Reaction 1**:
\[ \text{NH}_3(g) + 3\text{Cl}_2(g) \rightarrow \text{NCl}_3(g) + 3\text{HCl}(g) \quad \Delta H = \Delta H_1 \]

2. **Reaction 2**:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \quad \Delta H = \Delta H_2 \]

3. **Reaction 3**:
\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \quad \Delta H = \Delta H_3 \]

### Step 2: Analyze Reaction 1 to express ΔH1 in terms of heat of formation.

From Reaction 1, we can express ΔH1 as:
\[ \Delta H_1 = \text{(Heat of formation of NCl}_3) + 3 \times \text{(Heat of formation of HCl)} - \text{(Heat of formation of NH}_3) \]

### Step 3: Set the heat of formation of elements to zero.

The heat of formation of elements in their standard state is zero. Thus, for Cl2 and N2:
- Heat of formation of Cl2 = 0
- Heat of formation of N2 = 0

This simplifies our expression for ΔH1:
\[ \Delta H_1 = \Delta H_f(\text{NCl}_3) + 3 \Delta H_f(\text{HCl}) - \Delta H_f(\text{NH}_3) \]

### Step 4: Express the heat of formation of NH3 and HCl using ΔH2 and ΔH3.

From Reaction 2:
\[ \Delta H_2 = 2 \Delta H_f(\text{NH}_3) \implies \Delta H_f(\text{NH}_3) = \frac{\Delta H_2}{2} \]

From Reaction 3:
\[ \Delta H_3 = 2 \Delta H_f(\text{HCl}) \implies \Delta H_f(\text{HCl}) = \frac{\Delta H_3}{2} \]

### Step 5: Substitute the expressions for ΔH_f(NH3) and ΔH_f(HCl) back into the ΔH1 equation.

Substituting these into the ΔH1 equation gives:
\[ \Delta H_1 = \Delta H_f(\text{NCl}_3) + 3 \left(\frac{\Delta H_3}{2}\right) - \left(\frac{\Delta H_2}{2}\right) \]

### Step 6: Rearranging to solve for ΔH_f(NCl3).

Rearranging the equation to isolate ΔH_f(NCl3):
\[ \Delta H_f(\text{NCl}_3) = \Delta H_1 - \frac{3}{2} \Delta H_3 + \frac{1}{2} \Delta H_2 \]

### Final Expression:

Thus, the heat of formation of NCl3(g) in terms of ΔH1, ΔH2, and ΔH3 is:
\[ \Delta H_f(\text{NCl}_3) = \Delta H_1 - \frac{3}{2} \Delta H_3 + \frac{1}{2} \Delta H_2 \]