At 450 K, K_p = 2.0 xx 10^10/bar for the given reaction at equilibrium: 2SO_2(g) + O_2(g) hArr 2SO_3(g) What is the K_c at this temperature ?

At 450 K, K_p = 2.0 xx 10^10/bar for the given reaction at equilibrium

1.	At 450 K, K_p = 2.0 xx 10^10/bar for the given reaction at equilibrium: 2SO_2(g) + O_2(g) hArr 2SO_3(g) What is the K_c at this temperature ?
Answer» SOLUTION :`K_p=K_c(RT)^(Deltan_g)` or `K_c=K_p/(RT)^(Deltan_g)` `Deltan_g` =2-(2+1)=-1, T=450 K, R=0.083 L bar `K^(-1) "MOL"^(-1)` `THEREFORE K_c=(2.0xx10^10)/(0.083xx450)^(-1) =2.0xx10^10xx(0.083xx450)` `=7.47xx10^11 M^(-1)`