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I am dealing with the dissociation of water into its components hydrogen and oxygen and I am trying to determine the standard Gibbs free energies of each component. My professor said if the substance is an ideal gas in the standard state then

$$g_{i}^{0}=g_{i}^{\text{pure IG}}(T, P_\text{ref})$$

Now this means I can determine it for hydrogen and oxygen because they can be treated as ideal gases, but how would I consider the standard Gibbs free energy for water?

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Let's look at the reaction of the formation of water:

$\ce{2H2(g) + O2(g) -> 2H2O (l) }$

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 < 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}$.

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

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  • $\begingroup$ @BenNorris Thank you very much. I forgot how to do the dots and was having trouble with the exponents. $\endgroup$ – Jun-Goo Kwak Feb 28 '14 at 13:34

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