At 450 K, K_p = 2.0xx10^10/ bar for the given reaction of equilibrium . 2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) What is K_c at this temperature ?

At 450 K, K_p = 2.0xx10^10/ bar for the given reaction of equilibrium

1.	At 450 K, K_p = 2.0xx10^10/ bar for the given reaction of equilibrium . 2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) What is K_c at this temperature ?
Answer» Solution :`{:("Reaction at equilibrium:", 2SO_(2(G))+ , O_(2(g)) hArr , 2SO_(3(g))),("Stoichiometry of reaction:","2 mol","1 mol","2 mol"):}` `Deltan_((g))`= Total moles of product - Total mole of reactant =2 mol `(SO_3)` - (2 mol `SO_2` + 1 mol `O_2` ) =2-(2+1)=-1MOL `K_p=K_c(RT)^(Deltan)` and `K_c=K_p/(RT)^(Deltan)` `therefore K_c=(2.0xx10^10 "bar"^(-1))/([(0.0831 "bar L mol"^(-1) K^(-1))(450 K^(-1))])` `=(2.0xx10^10 "bar"^(-1))(0.0831 "bar L mol"^(-1) K^(-1))(450 K)` `=7.479xx10^11 "mol"^(-1) L = 7.479xx10^11 M^(-1)`