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. 0.088
B. 0.0533
C. 0.133
D. 0.177

Answer

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

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

we can use the equilibrium pressures given in the problem. The equilibrium pressures are:

- \( P_{\text{CO}_2} = 0.6 \) atm
- \( P_{\text{CO}} = 0.4 \) atm
- \( P_{\text{O}_2} = 0.2 \) atm

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

The expression for \( K_p \) in terms of partial pressures for the reaction is given by:

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

### Step 2: Substitute the equilibrium pressures into the expression

Now, substituting the values of the partial pressures into the expression:

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

### Step 3: Calculate the numerator and denominator

Calculating the numerator:

\[ (0.4)^2 = 0.16 \]

So,

\[ 0.16 \cdot 0.2 = 0.032 \]

Now calculating the denominator:

\[ (0.6)^2 = 0.36 \]

### Step 4: Divide the numerator by the denominator

Now, we can calculate \( K_p \):

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

### Step 5: Perform the division

Calculating the value:

\[ K_p = 0.08889 \approx 0.088 \]

### Final Answer

Thus, the value of \( K_p \) for the reaction at 3000 K is:

\[ K_p \approx 0.088 \]