The equilibrium constant for the reaction
`CO_(2)(g) +H_(2)(g) hArr CO(g) +H_(2)O(g) at 298 K` is `73`. Calculate the value of the standard free enegry change `(R =8.314 J K^(-1)mol^(-1))`

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

Equilibrium constant for the reaction: H_(2)(g) +I_(2) (g) hArr 2HI(g) is K_(c) = 50 at 25^(@)C The standard Gibbs free enegry change for the reaction will be:

Equilibrium constant for the reaction: H_(2)(g) +I_(2) (g) hArr 2HI(g) is K_(c) = 50 at 25^(@)C The standard Gibbs free enegry change for the reaction will be:

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

The value of equilibrium constant K_p for the reaction CO(g)+H_2O(g) leftrightarrow CO_2(g)+H_2(g) is 1.06 times 10^5 at 25^@C Calculate standard free energy change of the reaction at the same temperature.