A sample of pure `PCl_(5)` was introduced into an evacuted vessel at `473 K`. After equilibrium was attained,concentration of `PCl_(5)` was found to be `0.5xx10^(-1)mol litre^(-1)`. If value of `K_(c)` is `8.3xx10^(-3) mol litre^(-1)`. What are the concentrations of `PCl_(3)` and `Cl_(2)` at equilibrium ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium reaction can be represented as:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

### Step 2: Define initial concentrations and changes
Let the initial concentration of \( \text{PCl}_5 \) be \( A \) (which is unknown) and the initial concentrations of \( \text{PCl}_3 \) and \( \text{Cl}_2 \) be 0.

At equilibrium, the concentration of \( \text{PCl}_5 \) is given as:
\[ [\text{PCl}_5] = 0.05 \, \text{mol/L} \]

Let \( x \) be the change in concentration of \( \text{PCl}_5 \) that dissociates into \( \text{PCl}_3 \) and \( \text{Cl}_2 \). Therefore, at equilibrium:
- The concentration of \( \text{PCl}_5 \) will be \( A - x = 0.05 \)
- The concentration of \( \text{PCl}_3 \) will be \( x \)
- The concentration of \( \text{Cl}_2 \) will also be \( x \)

### Step 3: Relate \( A \) and \( x \)
From the equilibrium expression, we know:
\[ A - x = 0.05 \]

### Step 4: Substitute into the equilibrium expression
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{x \cdot x}{0.05} = \frac{x^2}{0.05} \]

Given that \( K_c = 8.3 \times 10^{-3} \, \text{mol/L} \), we can write:
\[ 8.3 \times 10^{-3} = \frac{x^2}{0.05} \]

### Step 5: Solve for \( x^2 \)
Rearranging gives:
\[ x^2 = 8.3 \times 10^{-3} \times 0.05 \]

Calculating \( x^2 \):
\[ x^2 = 4.15 \times 10^{-4} \]

### Step 6: Solve for \( x \)
Taking the square root to find \( x \):
\[ x = \sqrt{4.15 \times 10^{-4}} \]
\[ x \approx 2.03 \times 10^{-2} \, \text{mol/L} \]

### Step 7: Find concentrations of \( \text{PCl}_3 \) and \( \text{Cl}_2 \)
At equilibrium:
- \( [\text{PCl}_3] = x = 2.03 \times 10^{-2} \, \text{mol/L} \)
- \( [\text{Cl}_2] = x = 2.03 \times 10^{-2} \, \text{mol/L} \)

### Final Answer
The concentrations at equilibrium are:
- \( [\text{PCl}_3] = 2.03 \times 10^{-2} \, \text{mol/L} \)
- \( [\text{Cl}_2] = 2.03 \times 10^{-2} \, \text{mol/L} \)

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