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?

Answer

`7.47 xx 10^(11) M^(-1)`

Explanation

For the given reaction,
`Deltan = 2-3 = -1`
`T = 450 K`
`R = 0.0831 "bar" L "bar" K^(-1) "mol"^(-1)`
`K_(P) = 2.0 xx 10^(10) "bar"^(-1)`
We know that,
`K_(P) = K_(c) (RT) Deltan`
`rArr 2.0 xx 10^(10) "bar"^(-1) = K_(c)(0.0831 L "bar"K^(-1) "mol"^(-1) xx 450 K)^(-1)`
`rArr K_(c) = (2.0 xx 10^(10) "bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))`
`= 2.0 xx 10^(10) "bar"^(-1) = K_(c) (0.0831L "bar" K^(-1) "mol"^(-1) xx 450K)`
`rArr K_(c) = (2.0 xx 10^(10)"bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))`
`= (2.0 xx 10^(10) "bar"^(-1))(0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)`
`= 74.79 xx 10^(10) L "mol"^(-1)`
`= 7.48 xx 10^(11) L "mol"^(-1)`
`= 7.48 xx 10^(11) M^(-1)`