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 find the pressure necessary to obtain 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:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 2: Determine initial and equilibrium moles
Initially, we have:
- 1 mole of \( PCl_5 \)
- 0 moles of \( PCl_3 \)
- 0 moles of \( Cl_2 \)

When \( PCl_5 \) dissociates 80%, it means:
- 80% of 1 mole of \( PCl_5 \) dissociates, which is \( 0.8 \) moles.

At equilibrium, we have:
- Moles of \( PCl_5 \) remaining = \( 1 - 0.8 = 0.2 \) moles
- Moles of \( PCl_3 \) formed = \( 0.8 \) moles
- Moles of \( Cl_2 \) formed = \( 0.8 \) moles

### Step 3: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{moles of } PCl_5 + \text{moles of } PCl_3 + \text{moles of } Cl_2 = 0.2 + 0.8 + 0.8 = 1.8 \text{ moles} \]

### Step 4: Write the expression for the equilibrium constant \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]

### Step 5: Express partial pressures in terms of total pressure \( P \)
The partial pressures can be expressed in terms of the total pressure \( P \) as follows:
- \( P_{PCl_3} = \frac{0.8}{1.8} \cdot P \)
- \( P_{Cl_2} = \frac{0.8}{1.8} \cdot P \)
- \( P_{PCl_5} = \frac{0.2}{1.8} \cdot P \)

### Step 6: Substitute the partial pressures into the \( K_p \) expression
Substituting these into the \( K_p \) expression gives:
\[ K_p = \frac{\left(\frac{0.8}{1.8} P\right) \cdot \left(\frac{0.8}{1.8} P\right)}{\frac{0.2}{1.8} P} \]

### Step 7: Simplify the equation
This simplifies to:
\[ K_p = \frac{(0.8 \cdot 0.8) P^2}{0.2} \cdot \frac{1.8}{1.8^2} \]
\[ K_p = \frac{0.64 P^2}{0.2} \cdot \frac{1}{1.8} = \frac{0.64 P^2}{0.36} = 1.77 P \]

### Step 8: Conclusion
Thus, the pressure necessary to obtain 80% dissociation of \( PCl_5 \) is numerically equal to:
\[ K_p = 1.77 P \]