`PQ_(2)` dissociates as
`PQ_2 (g)hArrPQ(g)+Q(g)`
The initial pressure of `PQ_2` is 600 mm Hg. At equilibrium, the total pressure is 800mm Hg. Calculate the value of `K_p`.

Options

A. `100`
B. `400`
C. `200`
D. `50`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( PQ_2 \) can be represented as:
\[ PQ_2 (g) \rightleftharpoons PQ (g) + Q (g) \]

### Step 2: Set up the initial conditions
We are given that the initial pressure of \( PQ_2 \) is 600 mm Hg. At the start (time \( t = 0 \)):
- Pressure of \( PQ_2 \) = 600 mm Hg
- Pressure of \( PQ \) = 0 mm Hg
- Pressure of \( Q \) = 0 mm Hg

### Step 3: Define the change in pressure at equilibrium
Let \( x \) be the amount of \( PQ_2 \) that dissociates. At equilibrium, the pressures will be:
- Pressure of \( PQ_2 \) = \( 600 - x \) mm Hg
- Pressure of \( PQ \) = \( x \) mm Hg
- Pressure of \( Q \) = \( x \) mm Hg

### Step 4: Write the expression for total pressure at equilibrium
The total pressure at equilibrium is given as 800 mm Hg. Therefore, 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 substitute \( x \) back into the expressions for the pressures at equilibrium:
- Pressure of \( PQ_2 \) = \( 600 - 200 = 400 \) mm Hg
- Pressure of \( PQ \) = \( 200 \) mm Hg
- Pressure of \( Q \) = \( 200 \) mm Hg

### Step 7: Write the expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{PQ} \cdot P_Q}{P_{PQ_2}} \]
Substituting the equilibrium pressures into the equation:
\[ K_p = \frac{(200) \cdot (200)}{400} \]

### Step 8: Calculate \( K_p \)
Now we can calculate \( K_p \):
\[ K_p = \frac{40000}{400} = 100 \]

### Final Answer
Thus, the value of \( K_p \) is:
\[ \boxed{100} \]