For the reaction:

$$ \mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})} $$

we need to find the equilibrium constant $$\mathrm{K}_{\mathrm{c}}$$ at 300 K given that $$\mathrm{K}_{\mathrm{p}} = 0.492 \text{ atm}$$.

The relationship between $$\mathrm{K}_{\mathrm{p}}$$ and $$\mathrm{K}_{\mathrm{c}}$$ is given by the following equation:

$$ \mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{c}} (RT)^{\Delta n} $$

where:

$$R = 0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$$ is the gas constant,

$$T = 300 \, \mathrm{K}$$ is the temperature,

$$\Delta n$$ is the change in moles of gas from reactants to products.

For the reaction given:

The number of moles of gaseous products is 2 (for $$2 \mathrm{NO}_2$$).

The number of moles of gaseous reactants is 1 (for $$\mathrm{N}_2 \mathrm{O}_4$$).

Thus, $$\Delta n = 2 - 1 = 1$$.

Substitute the known values into the equation:

$$ 0.492 = \mathrm{K}_{\mathrm{c}} (0.082 \times 300)^{1} $$

Solve for $$\mathrm{K}_{\mathrm{c}}$$:

$$ 0.492 = \mathrm{K}_{\mathrm{c}} \times 24.6 $$

$$ \mathrm{K}_{\mathrm{c}} = \frac{0.492}{24.6} $$

Calculate $$\mathrm{K}_{\mathrm{c}}$$:

$$ \mathrm{K}_{\mathrm{c}} = 0.02 $$

Thus, $$\mathrm{K}_{\mathrm{c}}$$ for the reaction at 300 K is:

$$ 2 \times 10^{-2} $$