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 solve the problem, we will follow these steps:

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

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

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

### Step 4: Write the equilibrium expressions
At equilibrium, the moles will be:
- Moles of \( SO_3 = 1 - 2x \)
- Moles of \( SO_2 = 2x \)
- Moles of \( O_2 = x \)

### Step 5: Use the information given in the problem
We know that at equilibrium, the vessel contains 0.6 moles of \( SO_2 \):
\[ 2x = 0.6 \]
From this, we can solve for \( x \):
\[ x = \frac{0.6}{2} = 0.3 \]

### Step 6: Calculate the moles of each species at equilibrium
Now we can find the moles of \( SO_3 \) and \( O_2 \) at equilibrium:
- Moles of \( SO_3 = 1 - 2(0.3) = 1 - 0.6 = 0.4 \)
- Moles of \( O_2 = 0.3 \)

### Step 7: Calculate the concentrations
Since the volume of the flask is 1 litre, the concentrations are:
- Concentration of \( SO_3 = \frac{0.4}{1} = 0.4 \, \text{mol/L} \)
- Concentration of \( SO_2 = \frac{0.6}{1} = 0.6 \, \text{mol/L} \)
- Concentration of \( O_2 = \frac{0.3}{1} = 0.3 \, \text{mol/L} \)

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

### Step 9: Substitute the concentrations into the equilibrium expression
Substituting the values we calculated:
\[ K_c = \frac{(0.6)^2 \times (0.3)}{(0.4)^2} \]

### Step 10: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{0.36 \times 0.3}{0.16} = \frac{0.108}{0.16} = 0.675 \]

### Final Answer
The value of the equilibrium constant \( K_c \) is **0.675**.

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