The heat of formation of `PCl_(5)(s)` from the following data will be : `2P(s)+3Cl_(2)(g)to 2PCl_(3)(l), Delta H = - 151.8` Kcal and `PCl_(3)(l)+Cl_(2)(g)to PCl_(5)(s) , Delta H =- 32.8` Kcal

Options

A. `-108.7 Kcal`
B. `+108.7 Kcal`
C. `-184.6 Kcal`
D. `+184.6 Kcal`

Answer

To find the heat of formation of \( PCl_5(s) \) from the given reactions, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step-by-Step Solution:

1. **Write down the given reactions and their enthalpy changes:**
- Reaction 1:
\[ 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \quad \Delta H_1 = -151.8 \text{ Kcal} \]
- Reaction 2:
\[ PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \quad \Delta H_2 = -32.8 \text{ Kcal} \]

2. **Manipulate the reactions to find the formation of \( PCl_5(s) \):**
- We need to express the formation of \( PCl_5(s) \) from its elements. To do this, we can use Reaction 1 and Reaction 2.
- First, we will multiply Reaction 2 by 2 to match the coefficients of \( PCl_3 \) from Reaction 1:
\[ 2PCl_3(l) + 2Cl_2(g) \rightarrow 2PCl_5(s) \quad \Delta H_2' = 2 \times (-32.8) = -65.6 \text{ Kcal} \]

3. **Add the modified reactions:**
- Now, we can add Reaction 1 and the modified Reaction 2:
\[ 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \quad \Delta H_1 = -151.8 \text{ Kcal} \]
\[ 2PCl_3(l) + 2Cl_2(g) \rightarrow 2PCl_5(s) \quad \Delta H_2' = -65.6 \text{ Kcal} \]

- When we add these two reactions, we will cancel out \( 2PCl_3(l) \) and \( 2Cl_2(g) \):
\[ 2P(s) + 3Cl_2(g) + 2PCl_3(l) + 2Cl_2(g) \rightarrow 2PCl_5(s) \]
This simplifies to:
\[ 2P(s) + 5Cl_2(g) \rightarrow 2PCl_5(s) \]

4. **Calculate the total enthalpy change:**
- The total enthalpy change for the formation of \( 2PCl_5(s) \) is:
\[ \Delta H = \Delta H_1 + \Delta H_2' = -151.8 + (-65.6) = -217.4 \text{ Kcal} \]

5. **Find the heat of formation for one mole of \( PCl_5(s) \):**
- Since the reaction gives us the formation of 2 moles of \( PCl_5(s) \), we divide by 2:
\[ \Delta H_f(PCl_5) = \frac{-217.4}{2} = -108.7 \text{ Kcal} \]

### Final Answer:
The heat of formation of \( PCl_5(s) \) is \( -108.7 \text{ Kcal} \).