The standard Gibbs energies `(Delta_(f)G^(Theta))` for the formation of `SO_(2)(g)` and `SO_(3)(g)` are `-300.0` and `-371.0 kJ mol^(-1)` at `300K`, respectively. Calculate `DeltaG` and equilibrium constant for the following reaction at `300K`:
`2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)`

Explanation

`2SO_(2)(g) + O_(2)(g) iff 2SO_(3)(g)`
`Delta_(r)G^(@) = 2 Delta_(f)G^(@)(SO_(3)) - 2Delta_(f)G^(@)(SO_(2))- Delta_(f)G^(@)(O_(2))`
` = 2(-371) -2(-300) - 0`
` = - 742 + 600 = -142 kJ mol^(-1)`
Now `log K = -(Delta_(r)G^(@))/(2.303 RT)`
`Delta_(r)G^(@) = -142 kJ mol^(-1), R = 8.314 xx 10^(-3) kJ mol^(-1)K^(-1)`, T = 300K
`log K = -(-142)/(2.303 xx 8.314 xx 10^(-3)xx 300) = 24.72`
`therefore` K = Antilog (24.72)
`= 5.248 xx 10^(24)`.