For the reaction `C(s) + CO_2(g) Leftrightarrow 2CO(g),` the partial pressures of `CO_2` and Co are respectively 4 and 8 atm. `K_p` for the reaction is

Options

A. 16
B. 2
C. 0.5
D. 4

Answer

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

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

we will follow these steps:

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

The equilibrium constant \( K_p \) for the reaction in terms of partial pressures 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:
- The partial pressure of \( CO_2 \) (denoted as \( P_{CO_2} \)) is 4 atm.
- The partial pressure of \( CO \) (denoted as \( P_{CO} \)) is 8 atm.

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

\[ K_p = \frac{(8 \, \text{atm})^2}{4 \, \text{atm}} \]

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

Now, we calculate the value:

\[ K_p = \frac{64 \, \text{atm}^2}{4 \, \text{atm}} = 16 \, \text{atm} \]

### Final Answer

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

\[ K_p = 16 \, \text{atm} \]

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