Calculate equilibrium constant for the reaction:
`2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g) at 25^(@)C`
Given: `Delta_(f)G^(Theta) SO_(3)(g) = - 371.1 kJ mol^(-1)`,
`Delta_(f)G^(Theta)SO_(2)(g) =- 300.2 kJ mol^(-1)`
and `R = 8.31 J K^(-1) mol^(-1)`

Explanation

Step I: Calculation of `DeltaG^(Theta)` for the reaction
`DeltaG^(Theta) = sum Delta_(f)G^(Theta) (p) - sum Delta_(f)G^(Theta) (r )`
`=[2mol xx Delta_(f)G^(Theta)SO_(3)(g)]`
`-[2mol xx Delta_(f)G^(Theta) SO_(2),(g) +Delta_(f)G^(Theta) O_(2)(g)]`
`=[2mol xx 371.1 kJ mol^(-1)]`
`-[2mol xx (-300.2 kJ mol^(-1))]`
`=- 742.2 +600.4 =- 141.8 kJ`
Step II: Calculation of equilibrium constant `(K)`
`DeltaG^(Theta) =- 2.303 RT log K`
`DeltaG^(Theta) = - 141.8 kJ =- 141800 J, T = 25 +273 = 298K`,
`R = 8.31 JK^(-1) mol^(-1)`
`logK =- (DeltaG^(@))/(2.303 RT)`
`=(-) ((-141800J))/(2.303xx(8.31JK^(-1)mol^(-1)xx298K)) = 24.864`
`K = Antilog (24.864) = 7.31 xx 10^(24)`