A vessel at `1000K` contains `CO_(2)(g)` at `2` atm pressure. When graphite is added the following equilibrium is established.
`CO_(2)(g)+C(s) hArr 2CO(g)`
the toal pressure at equilibrium is 3 atm. The value of `K_(p)` is

Explanation

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

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

Given the initial conditions and the equilibrium state, we can follow these steps:

### Step 1: Initial Conditions
The vessel contains \( CO_2 \) at a pressure of 2 atm and no \( CO \) is present initially. The pressure of the solid carbon (graphite) does not affect the equilibrium expression since solids are not included in the \( K_p \) expression.

### Step 2: Change in Pressure
Let \( x \) be the change in pressure of \( CO_2 \) that reacts to form \( CO \). At equilibrium, the pressures will be:

- Pressure of \( CO_2 \) at equilibrium: \( 2 - x \) atm
- Pressure of \( CO \) at equilibrium: \( 2x \) atm

### Step 3: Total Pressure at Equilibrium
We know that the total pressure at equilibrium is 3 atm:

\[ (2 - x) + 2x = 3 \]

### Step 4: Solve for \( x \)
Simplifying the equation:

\[ 2 - x + 2x = 3 \]
\[ 2 + x = 3 \]
\[ x = 1 \text{ atm} \]

### Step 5: Calculate Partial Pressures at Equilibrium
Now we can find the partial pressures at equilibrium:

- Pressure of \( CO_2 \): \( 2 - x = 2 - 1 = 1 \) atm
- Pressure of \( CO \): \( 2x = 2 \times 1 = 2 \) atm

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) for the reaction is given by:

\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]

Substituting the equilibrium pressures:

\[ K_p = \frac{(2)^2}{1} = \frac{4}{1} = 4 \]

### Final Answer
Thus, the value of \( K_p \) is:

\[ \boxed{4} \]