Determining standard Gibbs free energy for water?
Let's look at the reaction of the formation of water:

$\ce{2H2(g) + O2(g) -&gt; 2H2O (l) }$

The Gibb's Free Energy relates the spontaneity of various reactions by looking at the change in enthalpy, temperature, and entropy.

We can relate those by:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

If we look at our reaction, it's clear that this is an exothermic reaction. Bonds are being formed as hydrogen gas and oxygen gas form water. Moles of gas are decreased and a liquid is formed so entropy is decreasing as well. We know that the formation of water is a spontaneous reaction, so $\Delta G^\circ$ must be negative. Therefore, this reaction must be run at temperatures where the difference of the enthalpy and the product of the entropy and temperature must be &lt; 0.

The Gibbs free energy is also a state function, so looking at $\Delta G^\circ_f$, we can find $\Delta G^\circ_{rxn}$.

$$\Delta G^\circ_{rxn} = \Sigma \Delta G^\circ_{products} - \Sigma \Delta G^\circ_{reactants} $$

Both hydrogen gas and oxygen gas are in their standard states and have a Gibbs Free energy of 0. If you look at tabulated standard thermodynamic data at $25^\circ\ \text{C}$, $\ce{H2O (l)}$ has a Gibbs free energy of$ -237.13 \text{kJ}\cdot\text{mol}^{-1}$.

$$\Delta G^\circ_{rxn} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} - 0 \text{ kJ}\cdot\text{mol}^{-1} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} $$