In the reaction : `C(s)+CO2(g)⇌2CO(g)`, the equilibrium pressure is 6 atm. If 50% of CO2 reacts, then Kp of the reaction is

Answer

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

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

Given that the equilibrium pressure is 6 atm and that 50% of \( CO_2 \) reacts, we can follow these steps:

### Step 1: Define Initial Conditions
Let the initial pressure of \( CO_2 \) be \( P \) atm. Since \( C \) is a solid, it does not appear in the equilibrium expression.

### Step 2: Determine Change in Pressure
If 50% of \( CO_2 \) reacts, then the change in pressure for \( CO_2 \) will be:
\[ \Delta P = -0.5P \]
Thus, at equilibrium, the pressure of \( CO_2 \) will be:
\[ P_{CO_2} = P - 0.5P = 0.5P \]

### Step 3: Calculate Pressure of \( CO \)
According to the stoichiometry of the reaction, for every mole of \( CO_2 \) that reacts, 2 moles of \( CO \) are produced. Therefore, the pressure of \( CO \) at equilibrium will be:
\[ P_{CO} = 2 \times (0.5P) = P \]

### Step 4: Total Pressure at Equilibrium
The total pressure at equilibrium can be expressed as:
\[ P_{total} = P_{CO_2} + P_{CO} = 0.5P + P = 1.5P \]
We are given that the total equilibrium pressure is 6 atm:
\[ 1.5P = 6 \text{ atm} \]

### Step 5: Solve for Initial Pressure \( P \)
To find \( P \):
\[ P = \frac{6}{1.5} = 4 \text{ atm} \]

### Step 6: Calculate Equilibrium Pressures
Now we can calculate the equilibrium pressures:
- For \( CO_2 \):
\[ P_{CO_2} = 0.5P = 0.5 \times 4 = 2 \text{ atm} \]
- For \( CO \):
\[ P_{CO} = P = 4 \text{ atm} \]

### Step 7: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{(4)^2}{2} = \frac{16}{2} = 8 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ \boxed{8} \]