A vessel of one litre capacity containing 1 mole of `SO_(3)` is heated till a state of equilibrium is attained.
`2SO_(3(g))hArr 2SO_(2(g))+O_(2(g))`
At equilibrium, 0.6 moles of `SO_(2)` had formed. The value of equilibrium constant is

Options

A. 0.18
B. 0.36
C. 0.45
D. 0.68

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:
\[ 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \]

### Step 2: Set up the initial conditions
We know from the problem statement that:
- The initial moles of \(\text{SO}_3\) = 1 mole
- The initial moles of \(\text{SO}_2\) = 0 moles
- The initial moles of \(\text{O}_2\) = 0 moles

### Step 3: Determine the change in moles at equilibrium
At equilibrium, it is given that 0.6 moles of \(\text{SO}_2\) are formed. According to the stoichiometry of the reaction, for every 2 moles of \(\text{SO}_2\) produced, 2 moles of \(\text{SO}_3\) are consumed, and 1 mole of \(\text{O}_2\) is produced.

Thus, if 0.6 moles of \(\text{SO}_2\) are formed, then:
- Moles of \(\text{SO}_3\) consumed = 0.6 moles (since the ratio is 1:1)
- Moles of \(\text{O}_2\) produced = \( \frac{0.6}{2} = 0.3 \) moles

### Step 4: Calculate the equilibrium moles
Now, we can calculate the moles at equilibrium:
- Moles of \(\text{SO}_3\) at equilibrium = Initial moles - Moles consumed = \(1 - 0.6 = 0.4\) moles
- Moles of \(\text{SO}_2\) at equilibrium = 0.6 moles (given)
- Moles of \(\text{O}_2\) at equilibrium = 0.3 moles (calculated)

### Step 5: Calculate the concentrations at equilibrium
Since the volume of the vessel is 1 liter, the concentrations (in molarity) will be equal to the number of moles:
- Concentration of \(\text{SO}_3\) = \(0.4 \, \text{M}\)
- Concentration of \(\text{SO}_2\) = \(0.6 \, \text{M}\)
- Concentration of \(\text{O}_2\) = \(0.3 \, \text{M}\)

### Step 6: Write the expression for the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the values we found:
\[ K_c = \frac{(0.6)^2 \times (0.3)}{(0.4)^2} \]

### Step 8: Calculate \(K_c\)
Calculating the values:
\[ K_c = \frac{(0.36) \times (0.3)}{(0.16)} = \frac{0.108}{0.16} = 0.675 \]

### Step 9: Round to appropriate significant figures
Rounding \(0.675\) gives us \(0.68\).

### Final Answer
The value of the equilibrium constant \(K_c\) is \(0.68\).

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