At K, K_(p) = 2*0 xx 10 ^(10)//"bar" for the given reaction at equilibrium 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) What is K_(c) at this temperature ?

At K, K_(p) = 2*0 xx 10 ^(10)//"bar" for the given reaction at equilib

1.	At K, K_(p) = 2*0 xx 10 ^(10)//"bar" for the given reaction at equilibrium 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) What is K_(c) at this temperature ?
Answer» Solution :For the given reaction , ` Delta n_(g)= 2 -3 = -1 ` ` K_(p) = K_(c) (RT)^(Delta n) or K_(c) = K_(p) (RT)^(-Delta n)= K_(p) (RT) = ( 20 xx 10^(10)"bar"^(-1))(00831 L "bar" K^(-1) "MOL" ^(-1) ) ( 450 K)` ` = 7.48 xx 10^(10 )L"mol"^(-1) = 7*48 xx 10^(11) L "mol"^(-1)`