`XY_2` dissociates as
`XY_2 (g) hArr XY(g) +Y(g)`
Initial pressure `XY_2` is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain cosntant ,the value of `K_p` is

Options

A. 50
B. 100
C. 20
D. 400

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of \( XY_2 \) can be represented as:
\[ XY_2 (g) \rightleftharpoons XY (g) + Y (g) \]

### Step 2: Define initial conditions
- The initial pressure of \( XY_2 \) is given as 600 mm Hg.
- At the start, the pressures of \( XY \) and \( Y \) are both 0 mm Hg.

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of \( XY_2 \) that dissociates at equilibrium. Therefore, at equilibrium:
- The pressure of \( XY_2 \) will be \( 600 - x \) mm Hg.
- The pressure of \( XY \) will be \( x \) mm Hg.
- The pressure of \( Y \) will also be \( x \) mm Hg.

### Step 4: Write the total pressure at equilibrium
The total pressure at equilibrium is given as 800 mm Hg. Thus, we can write:
\[ (600 - x) + x + x = 800 \]
This simplifies to:
\[ 600 + x = 800 \]

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

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

### Step 7: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{XY} \cdot P_Y}{P_{XY_2}} \]
Substituting the equilibrium pressures we found:
\[ K_p = \frac{(200)(200)}{400} \]

### Step 8: Calculate \( K_p \)
Calculating \( K_p \):
\[ K_p = \frac{40000}{400} = 100 \]

### Final Answer
The value of \( K_p \) is **100**.

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