One mole of `SO_(3)` was placed in a litre reaction flask at a given temperature when the reaction equilibrium was established in the reaction.
`2SO_(3) hArr 2SO_(2)+O_(2)` the vessel was found to contain `0.6` mol of `SO_(2)`. The value of the equilibrium constant is

Options

A. `0.36`
B. `0.675`
C. `0.45`
D. `0.54`

Explanation

To find the equilibrium constant \( K_c \) for the reaction:

\[ 2SO_3 \rightleftharpoons 2SO_2 + O_2 \]

we start with the information given in the problem.

### Step 1: Understand the initial conditions

We have 1 mole of \( SO_3 \) in a 1-liter flask. Therefore, the initial concentration of \( SO_3 \) is:

\[ [SO_3]_{initial} = 1 \, \text{mol/L} \]

### Step 2: Set up the change in concentration

Let \( x \) be the amount of \( SO_3 \) that dissociates at equilibrium. According to the stoichiometry of the reaction:

- For every 2 moles of \( SO_3 \) that dissociate, 2 moles of \( SO_2 \) and 1 mole of \( O_2 \) are produced.
- Therefore, at equilibrium, the concentrations will be:

\[ [SO_3] = 1 - x \]
\[ [SO_2] = 2x \]
\[ [O_2] = \frac{x}{2} \]

### Step 3: Use the given information

We are told that at equilibrium, the concentration of \( SO_2 \) is 0.6 mol. Since \( [SO_2] = 2x \):

\[ 2x = 0.6 \]

From this, we can solve for \( x \):

\[ x = \frac{0.6}{2} = 0.3 \]

### Step 4: Calculate the equilibrium concentrations

Now we can find the equilibrium concentrations of all species:

- For \( SO_3 \):

\[ [SO_3] = 1 - x = 1 - 0.3 = 0.7 \, \text{mol/L} \]

- For \( SO_2 \):

\[ [SO_2] = 2x = 0.6 \, \text{mol/L} \quad \text{(as given)} \]

- For \( O_2 \):

\[ [O_2] = \frac{x}{2} = \frac{0.3}{2} = 0.15 \, \text{mol/L} \]

### Step 5: Write the expression for the equilibrium constant

The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2} \]

### Step 6: Substitute the equilibrium concentrations into the expression

Now substituting the values we found:

\[ K_c = \frac{(0.6)^2 \cdot (0.15)}{(0.7)^2} \]

Calculating each part:

- \( (0.6)^2 = 0.36 \)
- \( (0.7)^2 = 0.49 \)

Now substituting these values into the equation:

\[ K_c = \frac{0.36 \cdot 0.15}{0.49} = \frac{0.054}{0.49} \approx 0.1102 \]

### Final Answer

Thus, the value of the equilibrium constant \( K_c \) is approximately:

\[ K_c \approx 0.1102 \]