`XY_(2)` dissociates `XY_(2)(g) hArr XY(g)+Y(g)`. When the initial pressure of `XY_(2)` is `600` mm Hg, the total equilibrium pressure is `800` mm Hg. Calculate K for the reaction Assuming that the volume of the system remains unchanged.

Options

A. `50.0`
B. `100.0`
C. `166.6`
D. `400.0`

Explanation

To solve the problem of the dissociation of \( XY_2 \) into \( XY \) and \( Y \), we will follow these steps:

### Step 1: Write the Reaction
The dissociation reaction is given as:
\[ XY_2(g) \rightleftharpoons XY(g) + Y(g) \]

### Step 2: Set Up Initial and Equilibrium Pressures
- **Initial Pressure of \( XY_2 \)**: \( P_{XY_2} = 600 \) mm Hg
- **Initial Pressure of \( XY \)**: \( P_{XY} = 0 \) mm Hg
- **Initial Pressure of \( Y \)**: \( P_{Y} = 0 \) mm Hg

At equilibrium, let \( x \) be the pressure of \( XY \) and \( Y \) formed. The dissociation of \( XY_2 \) will reduce its pressure by \( x \).

### Step 3: Express Equilibrium Pressures
At equilibrium:
- Pressure of \( XY_2 \) = \( 600 - x \)
- Pressure of \( XY \) = \( x \)
- Pressure of \( Y \) = \( x \)

### Step 4: Total Equilibrium Pressure
The total pressure at equilibrium is given as:
\[ P_{total} = P_{XY_2} + P_{XY} + P_{Y} \]
Substituting the values:
\[ 800 = (600 - x) + x + x \]
This simplifies to:
\[ 800 = 600 + x \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x = 800 - 600 = 200 \text{ mm Hg} \]

### Step 6: Calculate Equilibrium Pressures
Now we can find the equilibrium pressures:
- Pressure of \( XY_2 \) at equilibrium:
\[ P_{XY_2} = 600 - x = 600 - 200 = 400 \text{ mm Hg} \]
- Pressure of \( XY \) at equilibrium:
\[ P_{XY} = x = 200 \text{ mm Hg} \]
- Pressure of \( Y \) at equilibrium:
\[ P_{Y} = x = 200 \text{ mm Hg} \]

### Step 7: Write the Expression for \( K \)
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{P_{XY} \cdot P_{Y}}{P_{XY_2}} \]

### Step 8: Substitute the Values into the Equation
Substituting the equilibrium pressures into the expression for \( K \):
\[ K = \frac{(200)(200)}{400} \]

### Step 9: Calculate \( K \)
Calculating the value:
\[ K = \frac{40000}{400} = 100 \]

### Final Answer
Thus, the equilibrium constant \( K \) for the reaction is:
\[ \boxed{100} \]

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