A gaseous compound `XY_2` dissociates as:
`XY_2(g) hArr XY(g) + Y(g)`
If the initial pressure is 500 mm of Hg and the total pressure at equilibrium is 700 mm of Hg, calculate `K_p` for the reaction.

Explanation

To calculate the equilibrium constant \( K_p \) for the dissociation of the gaseous compound \( XY_2 \), we can follow these steps:

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

### Step 2: Identify initial conditions
The initial pressure of \( XY_2 \) is given as:
\[ P_{initial} = 500 \, \text{mm Hg} \]
At the start, only \( XY_2 \) is present, so:
\[ P_{XY_2} = 500 \, \text{mm Hg}, \quad P_{XY} = 0, \quad P_{Y} = 0 \]

### Step 3: Identify equilibrium conditions
The total pressure at equilibrium is given as:
\[ P_{total} = 700 \, \text{mm Hg} \]

### Step 4: Determine the change in pressure
Let \( x \) be the change in pressure due to the dissociation of \( XY_2 \). According to the stoichiometry of the reaction:
- For every 1 mole of \( XY_2 \) that dissociates, 1 mole of \( XY \) and 1 mole of \( Y \) are produced.

Thus, at equilibrium:
- The pressure of \( XY_2 \) will be \( 500 - x \)
- The pressure of \( XY \) will be \( x \)
- The pressure of \( Y \) will also be \( x \)

The total pressure at equilibrium can be expressed as:
\[ P_{total} = P_{XY_2} + P_{XY} + P_{Y} = (500 - x) + x + x = 500 + x \]

Setting this equal to the total pressure at equilibrium:
\[ 500 + x = 700 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x = 700 - 500 = 200 \, \text{mm Hg} \]

### Step 6: Calculate equilibrium pressures
Now we can find the equilibrium pressures:
- \( P_{XY_2} = 500 - x = 500 - 200 = 300 \, \text{mm Hg} \)
- \( P_{XY} = x = 200 \, \text{mm Hg} \)
- \( P_{Y} = x = 200 \, \text{mm Hg} \)

### Step 7: Calculate \( K_p \)
The equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{P_{XY} \cdot P_{Y}}{P_{XY_2}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{(200) \cdot (200)}{300} \]
\[ K_p = \frac{40000}{300} = 133.33 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ K_p = 133.33 \]