In the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the equilibrium pressure is `12` atm. If `50%` of `CO_(2)` reacts, calculate `K_(p)`.

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction:

\[ C(s) + CO_2(g) \rightleftharpoons 2CO(g) \]

Given that the equilibrium pressure is \( 12 \, \text{atm} \) and \( 50\% \) of \( CO_2 \) reacts, we can follow these steps:

### Step 1: Define Initial Conditions
Let the initial pressure of \( CO_2 \) be \( P \). Since \( 50\% \) of \( CO_2 \) reacts, the change in pressure will be \( -\frac{P}{2} \).

### Step 2: Calculate Equilibrium Pressures
At equilibrium:
- The pressure of \( CO_2 \) will be:
\[ P_{CO_2} = P - \frac{P}{2} = \frac{P}{2} \]
- The pressure of \( CO \) will be formed from the \( CO_2 \) that reacted. Since \( 1 \, \text{mol} \) of \( CO_2 \) produces \( 2 \, \text{mol} \) of \( CO \):
\[ P_{CO} = 2 \times \left( \frac{P}{2} \right) = P \]

### Step 3: Total Equilibrium Pressure
The total equilibrium pressure is given as \( 12 \, \text{atm} \):
\[ P_{total} = P_{CO_2} + P_{CO} = \frac{P}{2} + P = \frac{3P}{2} \]
Setting this equal to \( 12 \, \text{atm} \):
\[ \frac{3P}{2} = 12 \]

### Step 4: Solve for \( P \)
To find \( P \), multiply both sides by \( 2 \):
\[ 3P = 24 \]
Now, divide by \( 3 \):
\[ P = 8 \, \text{atm} \]

### Step 5: Calculate Partial Pressures at Equilibrium
Now we can find the partial pressures:
- \( P_{CO_2} = \frac{P}{2} = \frac{8}{2} = 4 \, \text{atm} \)
- \( P_{CO} = P = 8 \, \text{atm} \)

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]
Substituting the values we found:
\[ K_p = \frac{(8)^2}{4} = \frac{64}{4} = 16 \]

### Final Answer
Thus, the value of \( K_p \) is:
\[ \boxed{16} \]