`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 will follow these steps:

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

### Step 2: Define initial conditions.
- Initial pressure of \( XY_2 \) = 600 mm Hg
- Initial pressure of \( XY \) = 0 mm Hg
- Initial pressure of \( Y \) = 0 mm Hg

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

### Step 4: Write the total pressure at equilibrium.
The total pressure at equilibrium is given as 800 mm Hg. Therefore:
\[ (600 - P) + P + P = 800 \]
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 equilibrium pressures.
Now we can find the equilibrium pressures:
- Pressure of \( XY_2 \) at equilibrium = \( 600 - P = 600 - 200 = 400 \) mm Hg
- Pressure of \( XY \) at equilibrium = \( P = 200 \) mm Hg
- Pressure of \( Y \) at equilibrium = \( P = 200 \) mm Hg

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

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

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

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