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I've seen that Gibbs free energy is defined as the maximum amount of work that a system can do. From this I gather, this is how much energy it will release - right?

But, I've also learnt that the heat released when I burnt ethanol was due to the enthalpy of the reaction.

Given that the formula for Gibbs free energy is $\Delta G=\Delta H - T\Delta S$, it is quite clear that a change in enthalpy is not a change in Gibbs free energy. Rather, the entropy seems to also be able to contribute work.

If the total energy released by a reaction is greater than just the heat energy released as enthalpy, what form does the energy released by entropy take? Is there an intuitive way to understand where this energy is/goes?

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    $\begingroup$ Heat. You feel heat in the form of kinetic energy of air around you. $\endgroup$
    – Todd Minehardt
    Jul 22 at 15:17
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Suppose you have an ideal gas in a container in mechanical and thermal equilibrium with its surroundings (same $T$ and $p$—say, 298 K and 1 atm). It can't do any work on the surroundings.

Now compress it isothermally to, say, 100 atm. Now it certainly can do expansion work on the surroundings. What change in thermodynamic state function accounts for its increased ability to do work?

Well, since the internal energy, $U$, of a sample of ideal gas depends solely on its temperature, its internal energy hasn't changed. So the additional capacity to do work has not resulted from an increase in $U$.

But its free energy is much higher. Since $\Delta U = 0$ for that isothermal compression, the increase in the system's free energy—and the corresponding increase in the its ability to do work—has resulted solely from a decrease in its entropy.

So yes, a change in entropy can be used to do work.

It's tricky to talk formally about free energies for the above system, since while it's constant $T$, it's neither constant $p$ nor constant $V$. Rather it's constant $p \times V$.

So let's consider some more easily characterized systems (ideal gases at constant $T$ and $V$ or constant $T$ and $p$):

In a closed system at constant $T$ and $V$, the maximum total work the system can do, which is the reversible work, is given by $\Delta A$, the Helmholtz free energy. Under those conditions, $\Delta A=\Delta U - T\Delta S$. If $\Delta U = 0$, the magnitude of the maximum work is given by $T\Delta S$.

And in a closed system at constant $T$ and $p$, the maximum total non-$pV$ work the system can do, which is the reversible non-$pV$ work, is given by $\Delta G$, the Gibbs free energy. Under those conditions, $\Delta G = \Delta H - T\Delta S$. If $\Delta H= 0$, the magnitude of the maximum non-$pV$ work is given by $T\Delta S$.

So you can see we can formally connect the work a system can do with the entropy change. This shouldn't be surprising, because nothing happens to a macroscopic system unless it's associated with a decrease in the entropy of the universe. That's the 2nd law of thermodynamics, which provides an arrow of time for the universe. Thus anytime work is done anywhere, it's always ultimately driven by an increase in entropy (not of the system, but of the universe).

An increase in the entropy of the universe is the driving force behind every macroscopic process.

At constant $T$ and $p$, the only reason a decrease in enthalpy (release of thermal energy) corresponds to the ability of a system to undergo a change is because it causes an increase in the entropy of the surroundings: $\Delta H_{sys} = -T\Delta S_{surr}$. Thus ultimately the abilty to undergo any process, including those in which the system does work, isn't about the change in the energy of the system. It's about the change in the entropy of the universe.

Let's make this more general (i.e., not restrict our discussion to ideal gases): In a closed system at at constant $T$, $V$, both of the following hold (I'm taking positive dw to be work done on the system):

$$dA = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$ $$dA = đw_{max}$$

Thus:

$$đw_{max} = -T dS_{univ}$$

And in a closed system at at constant $T$, $p$ [G only gives non-$pV$-work because, at constant $p$, the $pV$-work is subtracted from G because it's subtracted from the enthalpy: $H = U +pV$; at constant $p$: $dH = dU + pdV = dU - đw_{pV}$.]:

$$dG = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$ $$dG = đw_{non-pV, max}$$

Thus:

$$đw_{non-pV,max} = -T dS_{univ}$$

In sum, it's all about the entropy, baby! The change in the energy or enthalpy of the system is only a bookkeeping surrogate for what really matters—the change in the entropy of the universe (unless you're doing industrial chemistry, and need to know the heat flow for practical reasons, in which case you would want to know $\Delta H$ or $\Delta U$, for processes at constant $p$ or $V$, respectively).

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  • $\begingroup$ That's a very cool point. However, isn't Gibbs free energy the non expansive work done? If so, then considering say burning fuel, what form does this "entropy energy" take? $\endgroup$
    – John Hon
    Jul 23 at 1:53
  • $\begingroup$ If you consider free energy in terms of Helmholtz free energy, then it is okay since the change in pressure doesn't matter. $\endgroup$
    – M.L
    Jul 23 at 2:23
  • $\begingroup$ @JohnHon You're correct, I was being too sloppy. At constant T and p, the decrease in the Gibbs free energy of a closed system is the theoretical maximum non-pV work that can be done. $\endgroup$
    – theorist
    Jul 23 at 6:05
  • $\begingroup$ @M.L Yes but then you need to constrain the system to constant volume for a decrease in Helmholtz free energy of the system to correspond to an increase in the entropy of the universe. $\endgroup$
    – theorist
    Jul 23 at 6:06
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I've seen that Gibbs free energy is defined as the maximum amount of work that a system can do. From this I gather, this is how much energy it will release - right?

No, this conclusion is incorrect, the Gibbs energy is not how much energy a process releases. A system can exchange energy with the surroundings in the form of heat or work. The Gibbs energy is a measure of the maximum non-PV work a system can do (if the Gibbs energy is negative), or the minimum non-PV work that has to be done on the system to make the process go forward (if the Gibbs energy is positive). The Gibbs energy is practical to use for constant pressure processes (where we know what the PV-work is for a given process).

But, I've also learnt that the heat released when I burnt ethanol was due to the enthalpy of the reaction.

That is true for a constant-pressure process in the absence of non-PV work. In the presence of non-PV work, you would have to add that to the heat exchanged (with the proper conventions for the +/- signs of heat and work).

Given that the formula for Gibbs free energy is $\Delta G=\Delta H - T\Delta S$, it is quite clear that a change in enthalpy is not a change in Gibbs free energy. Rather, the entropy seems to also be able to contribute work.

Both enthalpy and entropy contributions could have positive or negative values. So it might contribute to the work the system can do, or the work you have to do on the system to make the process go in the forward direction.

If the total energy released by a reaction is greater than just the heat energy released as enthalpy, what form does the energy released by entropy take?

There is no energy "released by entropy". The entropy portion determines how much choice there is of running a process with respect to work vs heat. In some processes, the work you can extract is larger than the enthalpy provided by the process (the additional energy comes from cooling down the surroundings).

One simple example of having a choice how to run a process is discharging a battery (i.e. an electrochemical reaction). You can short the battery (in the system), giving off heat to the surrounding. Or you can have some wires into the surrounding, running a motor. In this case you let the reaction do some work (there will probably also be some heat released).

Is there an intuitive way to understand where this energy is/goes?

I have not found an intuitive way to understand any of thermodynamics. Its beauty is the high level of abstraction, which might make it one of the less intuitive topics. I get surprised everytime I set it aside and then think about it again. One way to deal with this is to have a couple of examples you are comfortable with and where you know the proper description of what is going on. Then, you can return to these as you are looking at a situation that is new to you (or one you have forgotten about).

[from the comments] Can you give a concrete example of this for a non-electrochemical cell? I've only been able to understand this in terms of electrochemical reactions. How about say, combustion?

For a typical combustion reaction (e.g. burning some olive oil in an oil lamp), there is no work, just heat. However, our bodies manage to capture the free energy (e.g. heart using oxidation of fats as source of Gibbs energy to beat) by running the same reaction in steps near equilibrium. So again, in one case the chemical reaction produces heat only, and in the other case it generates a mix of work and heat. For a combustion reaction, it is possible to capture some of the thermal energy in a steam engine; however, you need some cool water that gets heated up as well, so you can't transform 100% of the thermal energy into work.

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  • $\begingroup$ See also: chemistry.stackexchange.com/q/49145/72973 and chemistry.stackexchange.com/q/112900/72973 $\endgroup$ Jul 23 at 5:19
  • $\begingroup$ that's interesting. Thank you. My question is, when you say that we can "extract" the energy, what form is this energy in. Can you give a concrete example of this for a non-electrochemical cell? I've only been able to understand this in terms of electrochemical reactions. How about say, combustion? $\endgroup$
    – John Hon
    Jul 23 at 8:21
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    $\begingroup$ @JohnHon For a typical combustion reaction (e.g. burning some olive oil in an oil lamp), there is no work, just heat. However, our bodies manage to capture the free energy (e.g. heart using oxidation of fats as source of Gibbs energy to beat). So again, in one case the chemical reaction produces heat only, and in the other case it generates a mix of work and heat. For a combustion reaction, it is possible to capture some of the thermal energy in a steam engine; however, you need some cool water that gets heated up as well, so you can't transform all the thermal energy into work. $\endgroup$ Jul 23 at 10:00
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    $\begingroup$ your last comment on the cells absorbing the work into the other chemicals was brilliant - totally helped me understand. Thank you! $\endgroup$
    – John Hon
    Jul 27 at 4:36
  • $\begingroup$ @KarstenTheis Oops, sorry, I meant to open up my answer for editing and accidentally opened yours--sorry for the two trivial edits! $\endgroup$
    – theorist
    Sep 15 at 19:47

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