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 of the reaction.

### Given Reactions:
1. \( 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \) \(\Delta H_1 = -151.8 \, \text{kcal}\)
2. \( PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \) \(\Delta H_2 = -32.8 \, \text{kcal}\)

### Steps to Calculate the Heat of Formation of \( PCl_5(s) \):

**Step 1: Adjust the first reaction.**
- We need only 1 mole of \( PCl_3 \) for the second reaction, so we will divide the first reaction by 2.

\[ \frac{1}{2}(2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l)) \implies P(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(l) \]

- The new enthalpy change for this reaction will be:

\[ \Delta H_1' = \frac{-151.8 \, \text{kcal}}{2} = -75.9 \, \text{kcal} \]

**Step 2: Write the second reaction as is.**
- The second reaction is already in the desired form:

\[ PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \quad \Delta H_2 = -32.8 \, \text{kcal} \]

**Step 3: Add the adjusted reactions.**
- Now we can add the two reactions together:

\[ P(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(l) \quad \Delta H_1' = -75.9 \, \text{kcal} \]
\[ PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \quad \Delta H_2 = -32.8 \, \text{kcal} \]

When we add these reactions, \( PCl_3(l) \) cancels out:

\[ P(s) + \frac{3}{2}Cl_2(g) + PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \]

This simplifies to:

\[ P(s) + 2Cl_2(g) \rightarrow PCl_5(s) \]

**Step 4: Calculate the total enthalpy change.**
- The total enthalpy change is the sum of the individual enthalpy changes:

\[ \Delta H = \Delta H_1' + \Delta H_2 = -75.9 \, \text{kcal} + (-32.8 \, \text{kcal}) = -108.7 \, \text{kcal} \]

### Conclusion:
The heat of formation of \( PCl_5(s) \) is \( -108.7 \, \text{kcal} \).