`AB_(2)` dissociates as
`AB_(2)(g) hArr AB(g)+B(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 solve the problem of calculating \( K_p \) for the dissociation of \( AB_2 \) into \( AB \) and \( B \), we can follow these steps:

### Step 1: Write the Reaction and Initial Conditions
The dissociation reaction is given as:
\[ AB_2(g) \rightleftharpoons AB(g) + B(g) \]

The initial pressure of \( AB_2 \) is given as:
\[ P_{AB_2, initial} = 500 \, \text{mm of Hg} \]

### Step 2: Define Changes at Equilibrium
Let \( x \) be the change in pressure due to the dissociation of \( AB_2 \). Therefore, at equilibrium:
- The pressure of \( AB_2 \) will be \( 500 - x \) mm of Hg.
- The pressure of \( AB \) will be \( x \) mm of Hg (since one mole of \( AB \) is produced for each mole of \( AB_2 \) that dissociates).
- The pressure of \( B \) will also be \( x \) mm of Hg (since one mole of \( B \) is produced for each mole of \( AB_2 \) that dissociates).

### Step 3: Write the Total Pressure at Equilibrium
The total pressure at equilibrium is given as:
\[ P_{total} = P_{AB_2} + P_{AB} + P_B \]
Substituting the pressures we defined:
\[ 700 = (500 - x) + x + x \]
This simplifies to:
\[ 700 = 500 + x \]
Thus, we find:
\[ x = 700 - 500 = 200 \, \text{mm of Hg} \]

### Step 4: Calculate the Equilibrium Pressures
Now we can calculate the equilibrium pressures:
- \( P_{AB_2} = 500 - x = 500 - 200 = 300 \, \text{mm of Hg} \)
- \( P_{AB} = x = 200 \, \text{mm of Hg} \)
- \( P_B = x = 200 \, \text{mm of Hg} \)

### Step 5: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{AB} \cdot P_B}{P_{AB_2}} \]

### Step 6: Substitute the Equilibrium Pressures into the Expression
Substituting the values we found:
\[ K_p = \frac{(200) \cdot (200)}{300} \]
\[ K_p = \frac{40000}{300} \]
\[ K_p = 133.33 \, \text{mm of Hg} \]

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

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