For the reaction $$SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g)$$, if $$K_P = K_C(RT)^x$$ where the symbols have usual meaning, then the value of x is: (assuming ideality)

For gaseous equilibria the general relation between the two equilibrium constants is first written:

$$K_P = K_C\,(RT)^{\Delta n}$$

where $$R$$ is the gas constant, $$T$$ is the absolute temperature, and $$\Delta n$$ denotes the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants.

The equilibrium under consideration is

$$SO_2(g) + \frac12\,O_2(g) \;\rightleftharpoons\; SO_3(g)$$

Now we evaluate $$\Delta n$$ step by step.

On the product side there is only one gaseous species, $$SO_3(g)$$, with a stoichiometric coefficient of $$1$$, so

$$n_{\text{products}} = 1$$

On the reactant side we have two gaseous species: $$SO_2(g)$$ with coefficient $$1$$ and $$O_2(g)$$ with coefficient $$\frac12$$. Therefore

$$n_{\text{reactants}} = 1 + \frac12 = \frac32$$

The change in the number of moles is thus

$$\Delta n = n_{\text{products}} - n_{\text{reactants}} = 1 - \frac32 = -\frac12$$

Substituting this value into the general formula gives

$$K_P = K_C\,(RT)^{-\,\frac12}$$

Comparing this with the form provided in the question, $$K_P = K_C\,(RT)^x$$, we clearly identify

$$x = -\frac12$$

Hence, the correct answer is Option B.