For the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the partial pressure of `CO_(2)` and `CO` is `2.0` and `4.0` atm, respectively, at equilibrium. The `K_(p)` of the reaction is

Options

A. `0.5`
B. `5.0`
C. `30.0`
D. `8.0`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

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

given the partial pressures of \( CO_2 \) and \( CO \) at equilibrium, we can follow these steps:

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction is given by the formula:

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

where \( P_{CO} \) is the partial pressure of carbon monoxide and \( P_{CO_2} \) is the partial pressure of carbon dioxide.

### Step 2: Substitute the given values

From the problem, we know:
- \( P_{CO_2} = 2.0 \, \text{atm} \)
- \( P_{CO} = 4.0 \, \text{atm} \)

Now we can substitute these values into the \( K_p \) expression:

\[ K_p = \frac{(4.0 \, \text{atm})^2}{2.0 \, \text{atm}} \]

### Step 3: Calculate \( K_p \)

Now, we calculate \( K_p \):

\[ K_p = \frac{16.0 \, \text{atm}^2}{2.0 \, \text{atm}} = 8.0 \, \text{atm} \]

### Final Answer

Thus, the value of \( K_p \) for the reaction is

\[ K_p = 8.0 \]

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