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Good day guys,

I am reading a book on electrochemical engineering, I came across some definitions of Gibbs energy and I am a bit confused as the book does not go into a lot of detail. The book first introduces Gibbs energy of a reaction $$\Delta G_{Rx} = \Delta G_f^{products} - \Delta G_f^{reactants}$$ and it mentions that $\Delta G_f$ is the gibbs energy of formation for a compound. And it refers me to a list in the appendix on Gibbs energy of formation values at ambient/standard conditions as $\Delta G$ and it does not mention pressures.

Later on in the book, the standard gibbs energy for a cell reaction is introduced as: $$ \Delta G^o_{Rx} = \sum_i s_i \Delta G_{f,i}^o $$ where $s_i$ are the stoichiometric coefficients and $\Delta G_{f,i}^o $ is the Gibbs energy of formation at standard/ambient conditions.

What exactly is the difference between $\Delta G^o$, $\Delta G_f $ and $\Delta G$ ? Does $\Delta G^o$ imply standard conditions and $\Delta G_f $ at some condition ?

Is $$\Delta G_{Rx} = \Delta H - T\Delta S = \sum_i s_i \Delta G_{f,i}^o $$ ?

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Here are the two definitions again:

$$\Delta G_{Rx} = \Delta G_f^{products} - \Delta G_f^{reactants}\tag{1}$$

$$ \Delta G^o_{Rx} = \sum_i s_i \Delta G_{f,i}^o \tag{2}$$

What exactly is the difference between $\Delta G^o$, $\Delta G_f $ and $\Delta G$ ? Does $\Delta G^o$ imply standard conditions and $\Delta G_f $ at some condition ?

The $^\circ$ added to a thermodynamic values does imply standard conditions. So you could rewrite equation 1 for standard conditions:

$$\Delta G^\circ_{Rx} = \Delta G{^\circ_f}^{products} - \Delta G{^\circ_f}^{reactants}\tag{1*}$$

Now, (1*) and (2) say exactly the same thing.

What is the difference between standard Gibbs energy and Gibbs energy of formation?

I will rephrase the question as "what is the difference between standard Gibbs energy of reaction and standard Gibbs energy of formation?". Typically, you would look up the standard Gibbs energy of formation, as those are the ones available. (Importantly, the standard state also sets partial pressures and concentrations, as the Gibbs energy is dependent on those). The symbols you show are for the Gibbs energy of reaction, a molar quantity (units Joule per mole). A change in Gibbs energy in general could also have dimensions of energy (units Joule), so it is important to differentiate.

So what is the difference?

The standard Gibbs energy of formation refers to a small set of standard Gibbs energies of reactions: Those with a single product and a coefficient of one, with reactants all elements in their allotrope present and most stable at standard conditions and the given temperature. This tells you that the standard Gibbs energy of formation for elements (the most stable allotrope in its physical state at the chosen temperature) is zero by definition.

Here is one such reaction:

$$\ce{1/8 S8(s) + 3/2 O2(g) -> SO3(g)}\tag{A}$$

For reaction (A), the standard Gibbs energy of reaction would be equal to the standard Gibbs energy of formation for sulfur trioxide.

I can write the reaction in a nicer form:

$$\ce{S8(s) + 12 O2(g) -> 8 SO3(g)}\tag{B}$$

For reaction (B), the standard Gibbs energy of reaction would be 8 times the standard Gibbs energy of formation of $\ce{SO3(g)}$, as equation (2) or (1*) will tell you.

We could have a different reaction that makes sulfur trioxide:

$$\ce{SO2(g) + 1/2 O2(g) -> SO3(g)}\tag{C}$$

To calculate the standard Gibbs energy of reaction (C), you would need to know the standard Gibbs energy of formation of $\ce{SO2(g)}$ as well.

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  • $\begingroup$ Just to make sure I understood correctly, the notation $\Delta G_f$ of products/reactants refers to the Gibbs energy of formation under non-standard conditions ? If standard conditions are used we denote this with the superscript (º) ? $\endgroup$
    – RMS
    Nov 27, 2023 at 23:00
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    $\begingroup$ Yes, or at least conditions that may include non-standard conditions. The definition is still valid if you happen to find yourself at standard conditions, of course, but also if not. I think in the textbook they were just lazy and did not include the superscript $^\circ$. $\endgroup$
    – Karsten
    Nov 27, 2023 at 23:51

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