For the reaction,
` CO(g) +(1)/(2) O_2(g) hArr CO_2 (g), `
` K_p//K_c ` is equal to

Options

A. 1
B. ` RT^(1//2)`
C. ` (RT)^(-(1)/(2))`
D. RT

Answer

To solve the problem of finding the relationship between \( K_p \) and \( K_c \) for the reaction:

\[ \text{CO(g)} + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \]

we will follow these steps:

### Step 1: Write the expressions for \( K_p \) and \( K_c \)

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

\[ K_p = \frac{P_{\text{CO}_2}}{P_{\text{CO}} \cdot P_{\text{O}_2}^{1/2}} \]

The equilibrium constant \( K_c \) in terms of concentrations is given by:

\[ K_c = \frac{[\text{CO}_2]}{[\text{CO}] \cdot [\text{O}_2]^{1/2}} \]

### Step 2: Relate partial pressures to concentrations

Using the ideal gas law, we know that:

\[ P = \frac{n}{V}RT \]

Thus, we can express the partial pressures in terms of concentrations:

\[ P_{\text{CO}} = [\text{CO}]RT, \quad P_{\text{O}_2} = [\text{O}_2]RT, \quad P_{\text{CO}_2} = [\text{CO}_2]RT \]

### Step 3: Substitute the expressions into \( K_p \)

Substituting these expressions into the equation for \( K_p \):

\[ K_p = \frac{[\text{CO}_2]RT}{[\text{CO}]RT \cdot ([\text{O}_2]RT)^{1/2}} \]

This simplifies to:

\[ K_p = \frac{[\text{CO}_2]RT}{[\text{CO}](RT)^{1/2} \cdot [\text{O}_2]^{1/2}(RT)^{1/2}} = \frac{[\text{CO}_2]}{[\text{CO}] \cdot [\text{O}_2]^{1/2}} \cdot \frac{RT}{(RT)^{1/2} \cdot (RT)^{1/2}} \]

### Step 4: Simplify the expression

Notice that \( (RT)^{1/2} \cdot (RT)^{1/2} = RT \):

\[ K_p = K_c \cdot \frac{RT}{RT} = K_c \cdot RT^{1/2} \]

### Step 5: Find the relationship \( \frac{K_p}{K_c} \)

Now, we can express the relationship between \( K_p \) and \( K_c \):

\[ \frac{K_p}{K_c} = RT^{1/2} \]

### Step 6: Final expression

For the given reaction, we find that:

\[ \frac{K_p}{K_c} = RT^{-1/2} \]

### Conclusion

Thus, the answer is:

\[ \frac{K_p}{K_c} = RT^{-\frac{1}{2}} \]