`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`
B. `100`
C. `200`
D. `400`

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Define initial conditions
- Initial pressure of \( XY_2 \) = 600 mm Hg
- At equilibrium, the total pressure = 800 mm Hg

### Step 3: Define changes at equilibrium
Let \( P \) be the pressure of \( XY_2 \) that dissociates at equilibrium. Therefore:
- Pressure of \( XY_2 \) at equilibrium = \( 600 - P \)
- Pressure of \( XY \) at equilibrium = \( P \)
- Pressure of \( Y \) at equilibrium = \( P \)

### Step 4: Write the equation for total pressure at equilibrium
The total pressure at equilibrium can be expressed as:
\[ (600 - P) + P + P = 800 \]
This simplifies to:
\[ 600 + P = 800 \]

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

### Step 6: Calculate the partial pressures at equilibrium
Now we can find the partial pressures:
- Partial pressure of \( XY_2 \) at equilibrium = \( 600 - P = 600 - 200 = 400 \, \text{mm Hg} \)
- Partial pressure of \( XY \) at equilibrium = \( P = 200 \, \text{mm Hg} \)
- Partial pressure of \( Y \) at equilibrium = \( P = 200 \, \text{mm Hg} \)

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

### Step 8: Substitute the values into the expression
Substituting the values we found:
\[ K_p = \frac{(200) \cdot (200)}{400} \]

### Step 9: Calculate \( K_p \)
Calculating gives:
\[ K_p = \frac{40000}{400} = 100 \]

### Conclusion
Thus, the equilibrium constant \( K \) for the reaction is:
\[ K = 100 \]