At 3000 K the equilibrium pressures of `CO_(2)` CO and `O_(2)` are `0.6, 0.4` and `0.2` atmospheres respectively. `K_(p)` fot the reaction, `2CO_(2)hArr2CO + O_(2)` is

Options

A. `0.089`
B. `0.0533`
C. `0.133`
D. `0.177`

Explanation

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

\[ 2 \text{CO}_2 \rightleftharpoons 2 \text{CO} + \text{O}_2 \]

given the equilibrium pressures of \( \text{CO}_2 \), \( \text{CO} \), and \( \text{O}_2 \) at 3000 K, we can follow these steps:

### Step 1: Write down the equilibrium pressures
The equilibrium pressures are given as follows:
- \( P_{\text{CO}_2} = 0.6 \) atm
- \( P_{\text{CO}} = 0.4 \) atm
- \( P_{\text{O}_2} = 0.2 \) atm

### Step 2: Write the expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = \frac{(P_{\text{CO}})^2 \cdot (P_{\text{O}_2})}{(P_{\text{CO}_2})^2} \]

### Step 3: Substitute the equilibrium pressures into the expression
Substituting the values into the expression:

\[ K_p = \frac{(0.4)^2 \cdot (0.2)}{(0.6)^2} \]

### Step 4: Calculate the values
Now, we calculate each part:

1. \( (0.4)^2 = 0.16 \)
2. \( (0.6)^2 = 0.36 \)

Now substituting these values back into the equation:

\[ K_p = \frac{0.16 \cdot 0.2}{0.36} \]

Calculating the numerator:

\[ 0.16 \cdot 0.2 = 0.032 \]

Now substituting this into the equation:

\[ K_p = \frac{0.032}{0.36} \]

### Step 5: Final calculation
Now, performing the division:

\[ K_p = 0.089 \]

Thus, the value of \( K_p \) for the reaction at 3000 K is approximately \( 0.089 \).

---