One mole of `SO_(3)` was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
The equilibrium mixture reacts with `0.2` mole `KMnO_(4)` in acidic medium. Hence, `K_(c)` is :

Options

A. `0.50`
B. `0.25`
C. `0.125`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

### Step 2: Set up the initial conditions
Initially, we have 1 mole of \( SO_3 \) in a 2-litre vessel. Therefore, the initial concentrations are:
- \( [SO_3] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \)
- \( [SO_2] = 0 \)
- \( [O_2] = 0 \)

### Step 3: Define the change in concentration at equilibrium
Let \( x \) be the amount of \( SO_3 \) that dissociates at equilibrium. According to the stoichiometry of the reaction:
- Change in \( SO_3 \): \( -2x \)
- Change in \( SO_2 \): \( +2x \)
- Change in \( O_2 \): \( +x \)

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [SO_3] = 0.5 - x \)
- \( [SO_2] = 2x \)
- \( [O_2] = x \)

### Step 5: Use the reaction with KMnO4 to find \( x \)
The problem states that the equilibrium mixture reacts with 0.2 moles of \( KMnO_4 \) in acidic medium. The reaction involves the oxidation of \( SO_2 \) to \( SO_3 \).

The equivalents of \( KMnO_4 \) can be calculated as:
\[ \text{Equivalents of } KMnO_4 = \text{moles} \times n \text{-factor} = 0.2 \times 5 = 1 \text{ equivalent} \]

For \( SO_2 \):
\[ \text{Equivalents of } SO_2 = \text{moles} \times n \text{-factor} = 2x \times 2 = 4x \text{ equivalents} \]

Setting the equivalents equal gives:
\[ 1 = 4x \]
\[ x = 0.25 \]

### Step 6: Calculate the equilibrium concentrations
Now substituting \( x \) back into the equilibrium expressions:
- \( [SO_3] = 0.5 - 0.25 = 0.25 \)
- \( [SO_2] = 2(0.25) = 0.5 \)
- \( [O_2] = 0.25 \)

### Step 7: Calculate the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.5)^2 (0.25)}{(0.25)^2} \]
Calculating this gives:
\[ K_c = \frac{0.25 \times 0.25}{0.0625} = \frac{0.0625}{0.0625} = 1 \]

### Final Result
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 1 \]