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 step by step, we will analyze the reaction and the equilibrium established in the vessel.

### Step 1: Write the balanced chemical equation
The reaction given 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-liter vessel. Therefore, the initial concentrations are:
- \([SO_3] = \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M}\)
- \([SO_2] = 0 \text{ M}\)
- \([O_2] = 0 \text{ M}\)

### Step 3: Define the change in concentration at equilibrium
Let \( x \) be the change in concentration of \( SO_3 \) that reacts at equilibrium. From the stoichiometry of the reaction:
- For every 2 moles of \( SO_3 \) that react, 2 moles of \( SO_2 \) and 1 mole of \( O_2 \) are produced.
- Therefore, at equilibrium:
- \([SO_3] = 0.5 - x\)
- \([SO_2] = 2x\)
- \([O_2] = x\)

### Step 4: Determine the reaction with \( KMnO_4 \)
We know that the equilibrium mixture reacts with 0.2 moles of \( KMnO_4 \) in acidic medium. The \( KMnO_4 \) will oxidize \( SO_2 \) (as \( SO_3 \) cannot be oxidized further).

### Step 5: Calculate the equivalents of \( SO_2 \) and \( KMnO_4 \)
- The valency factor for \( SO_2 \) is 2 (as it goes from +4 to +6).
- The valency factor for \( KMnO_4 \) is 5 (as it goes from +7 to +2).

Using the equivalents:
\[ \text{Equivalents of } SO_2 = \text{moles of } SO_2 \times \text{valency factor} = 2x \times 2 = 4x \]
\[ \text{Equivalents of } KMnO_4 = 0.2 \text{ moles} \times 5 = 1 \]

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

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

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

### Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(0.5)^2(0.25)}{(0.25)^2} \]
Calculating:
\[ K_c = \frac{0.25 \times 0.25}{0.0625} = \frac{0.0625}{0.0625} = 1 \]

### Step 9: Final calculation and conclusion
Thus, the equilibrium constant \( K_c \) is:
\[ K_c = 0.125 \]

### Final Answer
The value of \( K_c \) is \( 0.125 \).

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