For the reaction given below, the equilibrium constant K_c is 8.3 ×...

The given reaction is $2\mathrm{Na}_2\mathrm{O}_2(\mathrm{s}) + 2\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{Na}_2\mathrm{CO}_3(\mathrm{s}) + \mathrm{O}_2(\mathrm{g})$

The change in the number of moles of gaseous species is $\Delta n_g = n_p - n_r = 1 - 2 = -1$

The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$

Substituting the given values:

$K_p = 8.3 \times 10^3 \times (0.083 \times 500)^{-1}$

$K_p = \dfrac{8.3 \times 10^3}{41.5}$

$K_p = 200 = 2.0 \times 10^2$

Answer: $2.0 \times 10^2$