For the equilibrium reaction : 2H_(2)(g) + O_(2)(g) iff 2H_(2)O(l) at 298 K, DeltaG^(@) = - 474.78 kJ"mol"^(-1). Calculate log K for it. (R = 8.314 J K^(-1) "mol"^(-1)).

For the equilibrium reaction : 2H_(2)(g) + O_(2)(g) iff 2H_(2)O(l) at

1.	For the equilibrium reaction : 2H_(2)(g) + O_(2)(g) iff 2H_(2)O(l) at 298 K, DeltaG^(@) = - 474.78 kJ"mol"^(-1). Calculate log K for it. (R = 8.314 J K^(-1) "mol"^(-1)).
Answer» SOLUTION :`DeltaG^(@) = -2.303 RT"log K` or `log K = -(DeltaG^(@))/(2.303 RT)` `DeltaG^(@) = -474.78 kJ "mol"^(-1), R = 8.314 J K^(-1) "mol"^(-1)` T = 298 K `therefore log K = (-474.78 xx 10^(3)J"mol"^(-1))/(2..3030 xx 8.314 xx 298)` ` = 83.21` .