The ratio $$\frac{K_p}{K_c}$$ for the reaction $$CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g)$$ is:

The given reaction is:

$$CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g)$$

We need to find the ratio $$\frac{K_p}{K_c}$$. Recall that for any reaction, the relationship between the equilibrium constants in terms of partial pressures ($$K_p$$) and concentrations ($$K_c$$) is given by:

$$K_p = K_c (RT)^{\Delta n}$$

where $$R$$ is the universal gas constant, $$T$$ is the temperature in Kelvin, and $$\Delta n$$ is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

First, we calculate $$\Delta n$$ for the reaction. The reactants are $$CO(g)$$ and $$O_2(g)$$. The stoichiometric coefficients are:

• $$CO(g)$$: 1 mole
• $$O_2(g)$$: $$\frac{1}{2}$$ mole

So, the total moles of gaseous reactants = $$1 + \frac{1}{2} = \frac{3}{2}$$ moles.

The product is $$CO_2(g)$$:

• $$CO_2(g)$$: 1 mole

So, the total moles of gaseous products = 1 mole.

Therefore, the change in moles, $$\Delta n$$, is:

$$\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - \frac{3}{2} = -\frac{1}{2}$$

Now, substituting into the formula:

$$\frac{K_p}{K_c} = (RT)^{\Delta n} = (RT)^{-\frac{1}{2}}$$

We know that a negative exponent means taking the reciprocal, so:

$$(RT)^{-\frac{1}{2}} = \frac{1}{(RT)^{\frac{1}{2}}} = \frac{1}{\sqrt{RT}}$$

Therefore, the ratio $$\frac{K_p}{K_c} = \frac{1}{\sqrt{RT}}$$.

Comparing with the given options:

A. $$\frac{1}{\sqrt{RT}}$$

B. $$(RT)^{1/2}$$

C. RT

D. 1

We see that option A matches our result.

Hence, the correct answer is Option A.