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In the equation of Gibbs free energy change $(\Delta G) = (\Delta H)-T(\Delta S)$, is $T$ the system's or surrounding's temperature?

Edit: Oh sorry, I was not clear earlier, now I have a clear question. I know we have to calculate the Gibbs free energy of the system, but the criterion for spontaneity of $ΔS_\text{total}$ should be greater than zero. When we relate it to the Gibbs free energy to show that the Gibbs free energy change should be always negative, we keep both the system and surrounding temperature the same, and the pressure also constant. If so, how can the process occur? Wikipedia says that it is the chemical potential that undergoes changes there, but what about the Gibbs free energy change?

I think I was in too much of a hurry to ask a question when I was studying, but I am running out of time so I asked it here.

NEW EDIT TO QUESTION: I found that the above equation can only be applied to open systems, but the quoted statement was at the top of the Free energy of reactions section taken from Wikipedia Article

To derive the Gibbs free energy equation for an isolated system, let $S_\text{tot}$ be the total entropy of the isolated system, that is, a system that cannot exchange heat or mass with its surroundings. According to the second law of thermodynamics:

The word "isolated" system means that the formula $\Delta G=\Delta H-T\Delta S$ is derived for an isolated system. But, I think in exactly the opposite way, that it is true for open systems. Read the above question before this New EDIT to understand my question. Please Help me.

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    $\begingroup$ Why don't you answer your own question? You can do this and your "edit" is a good beginning. $\endgroup$ – WetSavannaAnimal Mar 30 '14 at 22:18
  • $\begingroup$ Sorry but I found my answer but just a last question so my thoughts then will become crystal clear. Question is -- Is it true that Work done by system other than PV work should always generate PV effect in surroundings? For eg:Electrical energy converted to EM wave that generate heating effect. $\endgroup$ – Vishvajeet Patil Mar 31 '14 at 13:56
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The temperature is that of both the system and surroundings. The whole point of "free energies" is that you must imagine:

  1. The system to be in thermodynamic equilibrium at all times so that, a fortiori this means that the system and environment must be at the same temperature, and,

  2. The system's thermodynamic macrostate to be "cycled", i.e., beginning and end temperatures must be the same; again a fortiori both environment and system are at the same temperature.

In my answer here I talk about the second point at length. (Also see user31748's answer where the words "... the environment must act as a thermostat that is either source or sink of the latent heat..." are a wonderful summary of the same ideas.) We talk of free energies as being the "available energy for work". Why is this something less than the enthalpy, or change in internal energies? Well actually, if you're allowed to "cheat" with your beginning and end temperatures and other thermodynamic macrostate variables, you CAN in principle actually extract all the enthalpy of an exothermic reaction as useful work - in my answer I show how this can be done with a system of cooled blocks that you can use to take your reactants down to a low temperature before allowing the reaction to happen (assuming it is still possible at near $0K$), thus extracting all the enthalpy as work and then use the same cold blocks in reverse to bring the reactants back up to their beginning temperature. The point is that you've changed the thermodynamic state of the blocks in a way that you can "get away" with your cheat for a small amount of reactants, but if you keep repeating the trick with more reactants, you will "degrade" the blocks and your "trick" won't work anymore. The ideas thus illustrate why a "reasonable" definition of free energy must keep everything at the same temperature - you can get all kinds of variations on the available work if your thermodynamic macrostate variables can change. You likely know that there is another "free energy" - the Helmholtz free energy - that makes another "reasonable" definition of beginning and end states - this time with the volume being constant (so you imagine your reaction happening inside an infinitely stiff, heat-permeable box).

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  • $\begingroup$ I just don't see anyone getting close to my question or just say approaching at the heart of question. Please read NEW EDIT in question $\endgroup$ – Vishvajeet Patil Mar 31 '14 at 12:51
  • $\begingroup$ Be very wary of the passage you cite: it uses the word "system" in two different ways. First it talks about an "isolated system" (being the only kind we can apply the second law $\Delta S >0$ to). Then it breaks this system up into the "internal" system and the surroundings; it quotes "...the closed system formed by both the system and its surroundings...". So its a bit sloppy. You clearly mean the reactants and products as the "system". Gibbs free energy applied to these when isolated doesn't make sense .... $\endgroup$ – WetSavannaAnimal Mar 31 '14 at 13:10
  • $\begingroup$ ... isolated reactants must in general change their temperature as the reaction goes forward - where else does generated heat go? $\endgroup$ – WetSavannaAnimal Mar 31 '14 at 13:11
  • $\begingroup$ @VishvajeetPatil See my comments above to you. $\endgroup$ – WetSavannaAnimal Mar 31 '14 at 13:30
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Think of the phase equilibrium of the vapor and liquid phases of a single component substance, say water. At constant pressure the temperature is also constant, (see Clausius - Clapeyron http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation ) and the equilibrium is described by the minimum of the Gibbs potential. The phase transition being reversible the temperature in question is both that of the environment and of the system. During phase transition the environment must act as a thermostat that is either source or sink of the latent heat of evaporation or of condensation, respectively.

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  • $\begingroup$ I like the words "environment must act as a thermostat". $\endgroup$ – WetSavannaAnimal Mar 30 '14 at 21:59
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It depends on what you want to calculate. For the Standard Free Energy, ΔG°, the temperature is 25°C. If you want to calculate the Free Energy at another temperature, ΔG, it is usually provided.

https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm

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  • $\begingroup$ You aren't clear. $\endgroup$ – Vishvajeet Patil Mar 30 '14 at 13:52
  • $\begingroup$ In essence, you choose the temperature you want to do the calculation at. $\endgroup$ – LDC3 Mar 30 '14 at 13:54
  • $\begingroup$ I just don't see anyone getting close to my question or just say approaching at the heart of question. Please read NEW EDIT in question. $\endgroup$ – Vishvajeet Patil Mar 31 '14 at 12:51
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I think you have made a wrong assumption:

gibbs free energy change should be always negative

That is not true. $\Delta G$ can be negative. The easiest way is to write the equation with the products as the reactants (most processes are reversible). Now the reaction is not spontaneous, but under the right conditions, (maybe increasing temperature), the products will convert to reactants.

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  • $\begingroup$ I already know that and I have no such assumption. $\endgroup$ – Vishvajeet Patil Mar 31 '14 at 14:17

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