`NH_(3)(g) + 3Cl_(2)(g) rarr NCl_(3)(g) + 3HCl(g), " "DeltaH_(1)`
`N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g), " "DeltaH_(2)`
`H_(2)(g)+ Cl_(2)(g) rarr 2HCl(g) , " " DeltaH_(3)`
The heat of formation of `NCl_(3)` 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

Explanation

To find the heat of formation of NCl₃ in terms of ΔH₁, ΔH₂, and ΔH₃, we will manipulate the given reactions. Here’s the step-by-step solution:

### Step 1: Write down the given reactions

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: Manipulate the reactions to find the desired formation reaction

We want to derive the formation of NCl₃ from the above reactions.

- From **Reaction 2**, we can divide the equation by 2:
\[ \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H = \frac{1}{2} \Delta H_2 \]

- From **Reaction 3**, we can reverse the equation and divide by 2:
\[ 3\text{HCl}(g) \rightarrow \frac{3}{2}\text{H}_2(g) + \frac{3}{2}\text{Cl}_2(g) \quad \Delta H = -\frac{3}{2} \Delta H_3 \]

### Step 3: Combine the reactions

Now we can add the modified **Reaction 2** and the reversed **Reaction 3** to **Reaction 1**:

1. **Add 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. **Add modified Reaction 2**:
\[ \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H = \frac{1}{2} \Delta H_2 \]

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

### Step 4: Write the overall reaction

When we add these reactions, we find:
\[ \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{Cl}_2(g) \rightarrow \text{NCl}_3(g) \]

### Step 5: Calculate the overall ΔH

The overall enthalpy change for the formation of NCl₃ is:
\[ \Delta H_f = \Delta H_1 + \frac{1}{2} \Delta H_2 - \frac{3}{2} \Delta H_3 \]

### Final Expression

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