A 1-litre container contains 2 moles of `PCI_5` initially. If at equilibrium, `K_c` is found to be 1, the degree of dissociation of `PCI_5` is

Options

A. 1
B. `-1`
C. `1/2`
D. 50

Answer

To solve the problem, we need to determine the degree of dissociation of \( \text{PCl}_5 \) in a 1-litre container where initially there are 2 moles of \( \text{PCl}_5 \). We also know that the equilibrium constant \( K_c \) is 1.

### Step-by-Step Solution:

1. **Write the Dissociation Reaction:**
The dissociation of \( \text{PCl}_5 \) can be represented as:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

2. **Set Up Initial Concentrations:**
Initially, we have:
- \( [\text{PCl}_5] = 2 \) moles in 1 litre, so \( [\text{PCl}_5] = 2 \, \text{M} \)
- \( [\text{PCl}_3] = 0 \)
- \( [\text{Cl}_2] = 0 \)

3. **Define the Change in Concentration:**
Let \( x \) be the degree of dissociation of \( \text{PCl}_5 \). At equilibrium:
- \( [\text{PCl}_5] = 2 - x \)
- \( [\text{PCl}_3] = x \)
- \( [\text{Cl}_2] = x \)

4. **Write the Expression for \( K_c \):**
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{x \cdot x}{2 - x} = \frac{x^2}{2 - x} \]

5. **Set Up the Equation Using Given \( K_c \):**
Since \( K_c = 1 \):
\[ \frac{x^2}{2 - x} = 1 \]

6. **Solve for \( x \):**
Rearranging the equation:
\[ x^2 = 2 - x \]
Rearranging gives:
\[ x^2 + x - 2 = 0 \]
Factoring the quadratic:
\[ (x - 1)(x + 2) = 0 \]
This gives us two possible solutions:
\[ x = 1 \quad \text{or} \quad x = -2 \]
Since \( x \) cannot be negative, we have:
\[ x = 1 \]

7. **Calculate the Degree of Dissociation:**
The degree of dissociation \( \alpha \) is given by:
\[ \alpha = \frac{x}{\text{initial moles}} = \frac{1}{2} = 0.5 \]

### Final Answer:
The degree of dissociation of \( \text{PCl}_5 \) is \( 0.5 \) or \( 50\% \).

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