At `300^@C` the pressure necessary to obtained 80% dissociation of `PCl_5` is numerically equal to

Options

A. `K_P = 17.7p`
B. `K_P = 1.77p`
C. `K_P = 2.22p`
D. `K_P = 177p`

Answer

To solve the problem of determining the pressure necessary to achieve 80% dissociation of \( PCl_5 \) at \( 300^\circ C \), we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

### Step 2: Define the initial conditions
Assume we start with 1 mole of \( PCl_5 \) and no products:
- Initial moles of \( PCl_5 \) = 1
- Initial moles of \( PCl_3 \) = 0
- Initial moles of \( Cl_2 \) = 0

### Step 3: Determine the change in moles due to dissociation
If 80% of \( PCl_5 \) dissociates, the amount dissociated is:
\[ 0.8 \text{ moles of } PCl_5 \]
Thus, the remaining \( PCl_5 \) will be:
\[ 1 - 0.8 = 0.2 \text{ moles} \]

### Step 4: Calculate the moles of products formed
From the dissociation, for every mole of \( PCl_5 \) that dissociates, 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \) are formed. Therefore, the moles of products formed will be:
- Moles of \( PCl_3 \) = 0.8
- Moles of \( Cl_2 \) = 0.8

### Step 5: Calculate the total moles at equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = \text{moles of } PCl_5 + \text{moles of } PCl_3 + \text{moles of } Cl_2 \]
\[ = 0.2 + 0.8 + 0.8 = 1.8 \text{ moles} \]

### Step 6: Determine the partial pressures
Let the total pressure be \( P \). The partial pressures can be calculated as follows:
- Partial pressure of \( PCl_5 \):
\[ P_{PCl_5} = P \cdot \frac{0.2}{1.8} \]
- Partial pressure of \( PCl_3 \):
\[ P_{PCl_3} = P \cdot \frac{0.8}{1.8} \]
- Partial pressure of \( Cl_2 \):
\[ P_{Cl_2} = P \cdot \frac{0.8}{1.8} \]

### Step 7: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(P \cdot \frac{0.8}{1.8}\right) \cdot \left(P \cdot \frac{0.8}{1.8}\right)}{P \cdot \frac{0.2}{1.8}} \]

### Step 8: Simplify the expression
This simplifies to:
\[ K_p = \frac{(P^2 \cdot \frac{0.64}{(1.8)^2})}{P \cdot \frac{0.2}{1.8}} = \frac{P \cdot 0.64}{0.2} = 3.2P \]

### Step 9: Solve for \( P \)
Given that \( K_p \) is numerically equal to \( P \) at equilibrium, we set:
\[ K_p = P \]
Thus:
\[ 3.2P = P \implies P = 2.22 \text{ (approx)} \]

### Conclusion
Therefore, the pressure necessary to obtain 80% dissociation of \( PCl_5 \) at \( 300^\circ C \) is approximately \( 2.22 \).