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I was looking at the enthalpy change for water-splitting reaction:

$$ \Delta H^o_R = [\Delta H^0_{H_2(g)} +\frac{1}{2}\Delta H^0_{O_2(g)}]-\Delta H^0_{H_2O(l)} = \pu{285.83 kJ/mol}$$

According to the book "Thermal physics"by Schroeder; at constant T and P; if there are no other forms of work on the system besides compression/expansion, then $\Delta H^o_R = Q$. However, when there are other forms of work being done we then have $\Delta Gº_R \leq W_{other}$ and $\Delta H^o_R = Q + W_{other}$. The value for the Gibbs free energy in this case is $ \Delta Gº_R = \pu{237 kJ/mol} $. We can relate $\Delta G$ and $\Delta H$ by $\Delta G =\Delta H -T\Delta S$ .

My confusion arises first from reading that the Gibbs free energy is the work we need to drive the reaction, say electrical work. However, the enthalpy change shows that the energy required could be done via heat and/or another form of work is higher than that of the Gibbs free energy? My guess so far is that we can take some energy from the environment for "free", but what happens when we drive this reaction only via heat, such that $\Delta H = Q$, would $\Delta G = 0 $ ?

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    $\begingroup$ To get the properly formatted $\pu{285.83 kJ/mol}$ you can write $\pu{285.83 kJ/mol}$ or $\pu{285.83 kJ mol-1}$ for $\pu{285.83 kJ mol-1}$. $\endgroup$
    – Poutnik
    Sep 6, 2023 at 9:06
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    $\begingroup$ Be aware that for the given conditions -ΔG/T = ΔS_total. You need to convert some nonvolume work to heat to ensure ΔS_total >= 0 $\endgroup$
    – Poutnik
    Sep 6, 2023 at 9:42

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First of all WE are doing nothing except possibly mixing chemicals and perhaps lighting the match. The energies involved reside in the internal energies in the reactants, products and the environs.

Enthalpy is U + PV. It is not necessarily heat because heat change is not a state function. The utility of enthalpy is that it is a state function and can be measured under certain conditions. Those conditions are at constant pressure and no work except PV work. Under these conditions Q, heat exchange, becomes a state function. Q can be measured by the art of calorimetry. Standard values are calculated from heat capacities and latent heats involved using Hess's Law. H, U, [even S] are based on a total amount of reaction; all the reactants become products. This causes a problem because all reactions do not go to completion.

We pchem students are not the only ones that struggle with this; imagine how the pioneers had to struggle: no textbooks, lousy equipment, no standards, fighting established beliefs, not even knowing what matter and energy really were [i am not saying that we do know now]. They accomplished wonders in developing the Laws of Thermodynamics, laws that seem almost infallible. One big question remained reactions did not go to completion, the question of equilibrium.

Gibbs, Helmholz and many others realized that there were two competing or complementary processes in literally everything: the release of energy and the increase of entropy. The increase in entropy happened because of the release of energy. When the energy release slowed the entropy increase slowed until at equilibrium they were equal. At equilibrium forward and reverse reactions have equal rates. There are some reactants that will not react. This effectively means that this heat from the reaction cannot be removed from the reaction. This energy is quantified by TdS. The maximum amount of heat that can be turned into work is A = U - TS at constant volume or G = H -TS at constant pressure. The First Law establishes that work and heat are equivalent. The Second Law states that some of the heat must remain at equilibrium and cannot be converted into work [without changing the conditions].

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[OP] what happens when we drive this reaction only via heat, such that ΔH=Q, would ΔG=0?

The Gibbs energy is a state function, so no matter how we run the reaction, the Gibbs energy of reaction stays the same. It is still positive, and that means you still need non-PV work to make the reaction go forward. As you know, water is stable at room temperature (it does not split into the elements while cooling the surrounding). So that hypothetical path is not available (in that direction). In the other direction, you have a choice. You can let the gas mixture explode, or feed it into a fuel cell to generate electricity.

[OP] Understanding the Role of Gibbs Free energy and Enthalpy as the energy required to drive a reaction

Enthalpy and Gibbs free energy are both state functions. So these are not parameters you can play with to drive a reaction. What you can do to drive a reaction are things like putting in work or removing a product (or - similarly - couple the reaction with a second one). Finally, you have no control over heat flow in a quasi-isothermal situation (you can't magically "push" heat from the solvent into the reactants).

[OP] My guess so far is that we can take some energy from the environment for "free"

Yes, that is true.

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  • $\begingroup$ Thank you very much for your explanations! I checked with the book again, and they sort of mention, that a reaction with $\Delta H > 0$ will only take a certain amount of heat: ($T\Delta S$), and the rest will need to be done as non-PV work. I am a bit thrown off by considering a "limit" to the heat we can use to drive the reaction (I thought Q could supply all $\Delta H^0$). Or is my understanding going astray somewhere? Kind regards. $\endgroup$
    – RMS
    Sep 7, 2023 at 18:01

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