`0.1 mol` of `PCl_(5)` is vaporised in a litre vessel at `260^(@)C`. Calculate the concentration of `Cl_(2)` at equilibrium, if the equilibrium constant for the dissociation of `PCl_(5)` is `0.0414`.

Explanation

To solve the problem step by step, we will analyze the dissociation of phosphorus pentachloride (PCl₅) and apply the equilibrium constant expression.

### Step 1: Write the dissociation reaction
The dissociation of phosphorus pentachloride can be represented as:
\[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]

### Step 2: Set up the initial concentrations
Given that we have `0.1 mol` of `PCl₅` in a `1 L` vessel, the initial concentration of `PCl₅` is:
\[ [\text{PCl}_5]_{initial} = 0.1 \, \text{mol/L} \]
The initial concentrations of `PCl₃` and `Cl₂` are both `0 mol/L` since they are products of the dissociation.

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of `PCl₅` that dissociates at equilibrium. Then, at equilibrium, the concentrations will be:
- \[ [\text{PCl}_5]_{eq} = 0.1 - x \]
- \[ [\text{PCl}_3]_{eq} = x \]
- \[ [\text{Cl}_2]_{eq} = x \]

### Step 4: Write the equilibrium constant expression
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
Substituting the equilibrium concentrations into the expression:
\[ 0.0414 = \frac{x \cdot x}{0.1 - x} \]
This simplifies to:
\[ 0.0414 = \frac{x^2}{0.1 - x} \]

### Step 5: Rearrange to form a quadratic equation
Multiplying both sides by \( (0.1 - x) \) gives:
\[ 0.0414(0.1 - x) = x^2 \]
Expanding this:
\[ 0.00414 - 0.0414x = x^2 \]
Rearranging leads to:
\[ x^2 + 0.0414x - 0.00414 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 0.0414, c = -0.00414 \):
- Calculate \( b^2 - 4ac \):
\[ b^2 = (0.0414)^2 = 0.00171396 \]
\[ 4ac = 4 \cdot 1 \cdot (-0.00414) = -0.01656 \]
Thus,
\[ b^2 - 4ac = 0.00171396 + 0.01656 = 0.01827396 \]
Now, substituting into the quadratic formula:
\[ x = \frac{-0.0414 \pm \sqrt{0.01827396}}{2} \]
Calculating the square root:
\[ \sqrt{0.01827396} \approx 0.135 \]
Now substituting back:
\[ x = \frac{-0.0414 \pm 0.135}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{-0.0414 + 0.135}{2} = \frac{0.0936}{2} = 0.0468 \)
2. \( x = \frac{-0.0414 - 0.135}{2} = \frac{-0.1764}{2} = -0.0882 \) (not valid since concentration cannot be negative)

### Step 7: Conclusion
The valid solution is:
\[ x = 0.0468 \, \text{mol/L} \]
Thus, the concentration of \( Cl_2 \) at equilibrium is:
\[ [\text{Cl}_2]_{eq} = 0.0468 \, \text{mol/L} \]