6
$\begingroup$

$$\mathrm dG < V\,\mathrm dp - S\,\mathrm dT$$

For a process at constant pressure and temperature ($\mathrm dp = \mathrm dT = 0$), and in the absence of any non-pV work, $\mathrm dG < 0$. Since energy cannot be created or destroyed (and Gibbs free energy is a measure of energy) where does the energy go when this spontaneous change occurs? Likewise for internal energy: $\mathrm dU = T\,\mathrm dS - p\,\mathrm dV$?

Also, why is this even useful? I understand that it's easy to hold temperature and pressure constant in a lab but we only seem to measure enthalpy anyway using $\mathrm dH = C_p\,\mathrm dT$. If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) rather than Gibbs free energy. I suspect I am barking up the wrong tree and I am hoping that you'll be able to put me right; the only thought that I had was that the fact that Gibbs free energy muct decrease for a spontaneous change would help to predict whether a spontaneous change would occur when done in a lab - given that $p$ and $T$ will be constant; is this right?

$\endgroup$
  • $\begingroup$ The inequality holds for a spontaneous or irreversible change; it'll be an equality for equilibrium or reversible change. $\endgroup$ – Fred Senese Jan 14 '15 at 17:50
  • 2
    $\begingroup$ @A.K., at risk of sounding like a pedant, I just want to point out that variables such as $G$, $V$, $p$, $T$, $S$ are supposed to be italicised. The only thing in the question that should be upright is the differential symbol $\mathrm d$. To put everything in roman (\mathrm or rm) is technically incorrect: there is a guide on meta, see e.g. chemistry.meta.stackexchange.com/q/443/16683 $\endgroup$ – orthocresol Sep 6 '18 at 14:32
7
$\begingroup$

If you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy).

If you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe.

When we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.


Let me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:

The Gibbs free energy is an energy in the sense that it is the part of the energy that is "free" to do useful work. It does indeed have units of energy, and $\Delta G$ can be used to compute the maximum possible nonexpansion work.

It is not an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G_\text{system} + G_\text{surroundings}$ isn't constant in a process.

If you have an isothermal process, $\Delta G= \Delta H−T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.

$\endgroup$
  • $\begingroup$ Gibbs free energy is still a measure of energy though isn't it? (it's units are Joules). So if it decreases, surely the energy of the system decreases doesn't it? $\endgroup$ – RobChem Jan 14 '15 at 19:42
  • $\begingroup$ If you have an isothermal process, $\Delta G = \Delta H - T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD is giving you a concrete example of that. $\endgroup$ – Fred Senese Jan 14 '15 at 19:52
  • $\begingroup$ So, is it reasonable to say that the Gibbs free energy isn't really energy? $\endgroup$ – RobChem Jan 14 '15 at 19:53
  • 1
    $\begingroup$ Gibbs free energy isn't an energy in the sense that it's conserved. It's derived from entropy. It's quite possible that $G_{system} + G_{surroundings}$ aren't constant in a process. $\endgroup$ – Fred Senese Jan 14 '15 at 19:59
  • $\begingroup$ ...That's the essence of my answer, really, if you want to track energy changes, look at energies (or enthalpies); if you want to look at spontaneity or equilibrium, look at Gibbs free energies. $\endgroup$ – Fred Senese Jan 14 '15 at 20:06
5
$\begingroup$

Imagine you have a divided insulated container. On one side is helium gas and on the other side is neon gas. Both sides are uniformly at standard temperature and pressure.

You remove the divider.

What happens?

The gases mix, because there are greatly more microstates we would consider "mixed" compare to not mixed, statisitically.

Entropy (S) increased, and in the ideal gas approximation there is no change in temperature or pressure (but there is a significant change in entropy of the system).

Gibbs so-called "free energy" decreased solely because entropy increased.

No energy went anywhere.

$\endgroup$
  • $\begingroup$ Very interesting. It'd be nice to see similar examples involving just vibrational, rotational or electronic entropy. Do you know any website or book? $\endgroup$ – santimirandarp Sep 6 '18 at 23:37
3
$\begingroup$

I think the other answers here are pretty good, but it seems to me that the main point of confusion is over what it means for energy to be conserved.

You know that total energy is conserved, but you also know that it can take different forms. Gibbs Free energy is just one form of energy. When Gibbs free energy decreases, it goes somewhere - either as heat released by the system, or into increasing the entropy of the system. Those are really the only two options, because this is how Gibbs energy is defined.

For your other questions:

why is this even useful?

Your thoughts at the end are correct - it is useful because it lets us determine spontaneity, and also the amount of energy available to do "useful" work - useful in the sense that we can control its transformation to other forms of energy.

If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant)

We track enthalpy even though it is the thermodynamic potential for constant P & S systems because at constant T & P, in the absence of non-PV work, it is also equal to the heat produced or absorbed by the system.

In the lab that might not seem like a big deal, but if you are designing a steam engine, air conditioning system, or a chemical reactor, being able to predict the amount of heat that is absorbed or lost during a transition is very useful!

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.