For the reaction given below, the equilibrium constant K_c is 8.3 10^3 at 500 ~K . 2 Na ₂ O ₂( s ) + 2 CO ₂( g ) 2 Na ₂ CO ₃( s ) + O ₂( g ) What is the value of K_p for this reaction at the same temperature? (Given: R = 0.083 ~L~bar~K⁻¹~mol⁻¹ )

Step-by-step solution

The given reaction is 2 Na ₂ O ₂( s ) + 2 CO ₂( g ) 2 Na ₂ CO ₃( s ) + O ₂( g ) The change in the number of moles of gaseous species is 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)^ n_g Substituting the given values: K_p = 8.3 10^3 (0.083 500)⁻¹ K_p = 8.3 10^3 41.5 K_p = 200 = 2.0 10^2 Answer: 2.0 10^2