`1` mol of `XY(g)` and `0.2` mol of `Y(g)` are mixed in `1` L vessel. At equilibrium, `0.6` mol of `Y(g)` is present. The value of `K` for the reaction
`XY(g)hArrX(g)+Y(g)` is

Options

A. `0.04 mol L^(-1)`
B. `0.06 mol L^(-1)`
C. `0.36 mol L^(-1)`
D. `0.40 mol L^(-1)`

Explanation

To solve the problem step by step, we will analyze the given reaction and the information provided.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ XY(g) \rightleftharpoons X(g) + Y(g) \]

### Step 2: Set up the initial conditions
We are given:
- Initial moles of \( XY = 1 \) mol
- Initial moles of \( Y = 0.2 \) mol
- Initial moles of \( X = 0 \) mol (since it is not present initially)

### Step 3: Define the changes at equilibrium
Let \( \alpha \) be the amount of \( XY \) that dissociates at equilibrium. Therefore, at equilibrium:
- Moles of \( XY = 1 - \alpha \)
- Moles of \( X = \alpha \)
- Moles of \( Y = 0.2 + \alpha \)

### Step 4: Use the information given about equilibrium
We know from the problem that at equilibrium, the moles of \( Y \) is \( 0.6 \) mol. Therefore:
\[ 0.2 + \alpha = 0.6 \]

### Step 5: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha = 0.6 - 0.2 = 0.4 \]

### Step 6: Calculate the moles of each component at equilibrium
Now substituting \( \alpha \) back into our expressions for the moles:
- Moles of \( XY = 1 - 0.4 = 0.6 \)
- Moles of \( X = 0.4 \)
- Moles of \( Y = 0.6 \)

### Step 7: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[X][Y]}{[XY]} \]
Where:
- \([X] = \frac{0.4}{1} = 0.4 \, \text{mol/L}\)
- \([Y] = \frac{0.6}{1} = 0.6 \, \text{mol/L}\)
- \([XY] = \frac{0.6}{1} = 0.6 \, \text{mol/L}\)

### Step 8: Substitute the values into the \( K_c \) expression
Now substituting the values into the expression:
\[ K_c = \frac{(0.4)(0.6)}{0.6} \]

### Step 9: Simplify the expression
\[ K_c = \frac{0.24}{0.6} = 0.4 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is:
\[ K_c = 0.4 \, \text{mol/L} \]