The equilibrium constant `K_(p)` for the thermal dissociation of `PCl_(5)` at `200^(@)C` is 1.6 atm. The pressure (in atm) at which it is `50%` dissociated at that temperature is

Options

A. 4.8
B. 4.2
C. 3.2
D. 2.4

Answer

To solve the problem, we need to determine the pressure at which PCl₅ is 50% dissociated at 200°C, given that the equilibrium constant \( K_p \) is 1.6 atm.

### Step-by-Step Solution:

1. **Write the equilibrium reaction**:
The thermal dissociation of PCl₅ can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

2. **Define initial conditions**:
Let's assume we start with 1 mole of PCl₅. Therefore, the initial moles of PCl₅ are:
\[ [PCl_5]_{initial} = 1 \text{ mole} \]
and the initial moles of PCl₃ and Cl₂ are:
\[ [PCl_3]_{initial} = 0 \text{ moles}, \quad [Cl_2]_{initial} = 0 \text{ moles} \]

3. **Define the change at equilibrium**:
If PCl₅ is 50% dissociated, then:
\[ x = 0.5 \text{ moles of PCl}_5 \text{ dissociated} \]
Thus, at equilibrium:
\[ [PCl_5]_{equilibrium} = 1 - x = 0.5 \text{ moles} \]
\[ [PCl_3]_{equilibrium} = x = 0.5 \text{ moles} \]
\[ [Cl_2]_{equilibrium} = x = 0.5 \text{ moles} \]

4. **Calculate total moles at equilibrium**:
The total number of moles at equilibrium is:
\[ \text{Total moles} = [PCl_5] + [PCl_3] + [Cl_2] = 0.5 + 0.5 + 0.5 = 1.5 \text{ moles} \]

5. **Determine partial pressures**:
Assuming the total pressure is \( P \):
- The partial pressure of PCl₅:
\[ P_{PCl_5} = \frac{0.5}{1.5} \times P = \frac{P}{3} \]
- The partial pressure of PCl₃:
\[ P_{PCl_3} = \frac{0.5}{1.5} \times P = \frac{P}{3} \]
- The partial pressure of Cl₂:
\[ P_{Cl_2} = \frac{0.5}{1.5} \times P = \frac{P}{3} \]

6. **Write the expression for \( K_p \)**:
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{P}{3}\right) \cdot \left(\frac{P}{3}\right)}{\frac{P}{3}} = \frac{P^2/9}{P/3} = \frac{P}{3} \]

7. **Set \( K_p \) equal to the given value**:
We know \( K_p = 1.6 \) atm, so:
\[ \frac{P}{3} = 1.6 \]

8. **Solve for \( P \)**:
Multiplying both sides by 3:
\[ P = 1.6 \times 3 = 4.8 \text{ atm} \]

### Final Answer:
The pressure at which PCl₅ is 50% dissociated at 200°C is **4.8 atm**.