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

Explanation

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) = ( 2*0 xx 10^(10)"bar"^(-1))(0*0831 L "bar" K^(-1) "mol" ^(-1) ) ( 450 K)`
` = 7.48 xx 10^(10 )L"mol"^(-1) = 7*48 xx 10^(11) L "mol"^(-1)`