At 700 K the equilibrium constant K_(p) for the reaction 2 SO_(3) (g) hArr 2 SO_(2) (g) + O_(2) (g) is 1.80 xx10^(-3)kPa. What is the numerical value of K_(c) in moles per litre for this reaction at the same temperature ?

At 700 K the equilibrium constant K_(p) for the reaction 2 SO_(3) (g)

1.	At 700 K the equilibrium constant K_(p) for the reaction 2 SO_(3) (g) hArr 2 SO_(2) (g) + O_(2) (g) is 1.80 xx10^(-3)kPa. What is the numerical value of K_(c) in moles per litre for this reaction at the same temperature ?
Answer» Solution :Here `n_(p) =3 " moles" , n_(r) = 2 " moles "` `:. " "DELTAN = n_(p) - n _(r) = 3- 2 = 1 ` ` K_(p) = 1.80 xx 10^(-3) kPA = 1. 80 PA = (1.80)/(10^(5)) "bar" = 1.80 xx 106(-5) " bar"` `R= 0.0831 L " bar " K^(-1) mol^(-1), T=700 K` USING the relation , `K_(p) = K_(C) (RT)^Delta n` `K_(c) = (K_(p))/(RT)= (1.80 xx 10^(-5) "bar")/(0.0831 L " bar " K^(-1) xx 700 K )= 3.09 xx 106(-7) " mol " L^(-1)` `" Alternatively, " K_(c) = (K_(p))/(RT) =(1.8Pa)/((8.314 JK^(-1) mol^(-1))(700 K))""( :' Pa = N m^(-2), J = N m)` `= 3.09 xx 10^(-4)" mol " m^(-3) = 3.09 xx 10^(-7)" mol " dm^(-3) or" mol " L^(-1)`