For the equilibrium
`H_(2) O (1) hArr H_(2) O (g)`
at 1 atm `298 K`

For the equilibrium H_(2)O(l) iff H_(2)O(g) at 1 atm and 298 K

For the following equilibrium, H_(2) O(l) hArr H_(2) O(g) the increase in the pressure causes

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)) .

For the equilibrium 2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g) , equilibrium constant is K_(1) For the equilibrium 2CO_(2)hArr 2CO(g)+O_(2)(g) , equilibrium constant is K_(2) The equilibrium constant for : CO_(2)(g)+H_(2)(g)hArr CO(s)+H_(2)O(g) is :

For the equilibrium, 2H_(2)O hArr H_(3)O^(+) + OH^(-) ,the value of DeltaG^(@) at 298 K is approximately:

For the equilibrium, 2H_(2)O hArr H_(3)O^(+) + OH^(-) ,the value of DeltaG^(@) at 298 K is approximately:

For the equilibrium, 2H_(2)O hArr H_(3)O^(+) + OH^(-) ,the value of DeltaG^(@) at 298 K is approximately: