`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 step by step, we will analyze the dissociation of the gas \( AB_2 \) and use the information provided to calculate the equilibrium constant \( K_p \).

### Step 1: Write the balanced equation for the dissociation
The dissociation reaction is given as:
\[ AB_2(g) \rightleftharpoons AB(g) + B(g) \]

### Step 2: Define the initial conditions
The initial pressure of \( AB_2 \) is given as:
\[ P_{AB_2} = 500 \, \text{mm Hg} \]
At the start, the pressures of \( AB \) and \( B \) are both \( 0 \, \text{mm Hg} \).

### Step 3: Define the change in pressure at equilibrium
Let \( x \) be the amount of \( AB_2 \) that dissociates at equilibrium. Therefore, at equilibrium:
- The pressure of \( AB_2 \) will be:
\[ P_{AB_2} = 500 - x \]
- The pressure of \( AB \) will be:
\[ P_{AB} = x \]
- The pressure of \( B \) will also be:
\[ P_B = x \]

### Step 4: Write the expression for total pressure at equilibrium
The total pressure at equilibrium is given as:
\[ P_{total} = P_{AB_2} + P_{AB} + P_B \]
Substituting the expressions we have:
\[ P_{total} = (500 - x) + x + x = 500 + x \]

### Step 5: Set up the equation using the total pressure at equilibrium
We know from the problem that the total pressure at equilibrium is \( 700 \, \text{mm Hg} \):
\[ 500 + x = 700 \]

### Step 6: Solve for \( x \)
Rearranging the equation gives:
\[ x = 700 - 500 = 200 \, \text{mm Hg} \]

### Step 7: Calculate the partial pressures at equilibrium
Now we can find the partial pressures:
- For \( AB_2 \):
\[ P_{AB_2} = 500 - x = 500 - 200 = 300 \, \text{mm Hg} \]
- For \( AB \):
\[ P_{AB} = x = 200 \, \text{mm Hg} \]
- For \( B \):
\[ P_B = x = 200 \, \text{mm Hg} \]

### Step 8: 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 9: Substitute the values into the \( K_p \) expression
Substituting the values we found:
\[ K_p = \frac{(200) \cdot (200)}{300} \]

### Step 10: Calculate \( K_p \)
Calculating this gives:
\[ K_p = \frac{40000}{300} = 133.33 \, \text{mm Hg} \]

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

---