`2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)`
Starting with 2 mol of `SO_(2)` and 1 mol `O_(2)` in one litre flask. After some time equilibrium established. In another experiment same mol of `SO_(2)` presetn at equilibrium required 1000 ml 0.4 M `KMnO_(4)` ub acidic medium. The value of equilibrium constant `K_(C)` for the reaction is

Explanation

To solve the problem step by step, we will follow the chemical equilibrium concepts and calculations.

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

### Step 2: Set up the initial conditions
Initially, we have:
- 2 moles of \( SO_2 \)
- 1 mole of \( O_2 \)
- 0 moles of \( SO_3 \)

Since the volume of the flask is 1 liter, the initial concentrations are:
- \([SO_2] = 2 \, \text{M}\)
- \([O_2] = 1 \, \text{M}\)
- \([SO_3] = 0 \, \text{M}\)

### Step 3: Define the change in concentration at equilibrium
Let \( x \) be the change in moles of \( SO_2 \) that reacts to reach equilibrium. The changes in concentration will be:
- \( SO_2 \): \( 2 - x \)
- \( O_2 \): \( 1 - \frac{x}{2} \)
- \( SO_3 \): \( x \)

### Step 4: Use the information from the second experiment
In the second experiment, the amount of \( SO_2 \) present at equilibrium is determined using 1000 mL of 0.4 M \( KMnO_4 \) in acidic medium.

The equivalents of \( KMnO_4 \) used can be calculated as:
\[ \text{Equivalents of } KMnO_4 = \text{Volume (L)} \times \text{Molarity} \times \text{n} \]
Where \( n \) is the number of electrons transferred in the redox reaction. For \( KMnO_4 \) in acidic medium, \( n = 5 \).

Calculating:
\[ \text{Equivalents of } KMnO_4 = 1 \, \text{L} \times 0.4 \, \text{mol/L} \times 5 = 2 \, \text{equivalents} \]

### Step 5: Relate the equivalents of \( SO_2 \) to \( KMnO_4 \)
Since the equivalents of \( SO_2 \) are equal to the equivalents of \( KMnO_4 \), we have:
\[ \text{Equivalents of } SO_2 = \text{Moles of } SO_2 \times 2 = 2 \]
Thus,
\[ \text{Moles of } SO_2 = 1 \, \text{mol} \]

### Step 6: Substitute \( x \) into the equilibrium expression
From the equilibrium expression, we know:
\[ 2 - x = 1 \implies x = 1 \]

### Step 7: Calculate the equilibrium concentrations
Substituting \( x \) into the equilibrium concentrations:
- \([SO_2] = 2 - 1 = 1 \, \text{M}\)
- \([O_2] = 1 - \frac{1}{2} = 0.5 \, \text{M}\)
- \([SO_3] = x = 1 \, \text{M}\)

### Step 8: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(1)^2}{(1)^2(0.5)} = \frac{1}{0.5} = 2 \]

### Final Answer
The value of the equilibrium constant \( K_c \) for the reaction is:
\[ \boxed{2} \]