[orgo814]

I came across a true and false question saying "A positive entropy change will always be spontaneous". I believe this is true but that leads to another question. They say that negative enthalpy change, positive entropy change, and negative free energy change is always spontaneous. But, what if we had a really low temperature and utilizing the equation delta G= delta H- TdeltaS, we could have a positive delta H, positive delta S, and then a positive delta G. Therefore, the delta G is saying that the reaction is not spontaneous while the entropy change is saying it is. What value should I ultimately look at to determine whether or not this is a spontaneous reaction? Thank you.

[UG]

What value should I ultimately look at to determine whether or not this is a spontaneous reaction?

Ultimately still the change in Gibb's free energy.

[fledarmus]

They say that negative enthalpy change, positive entropy change, and negative free energy change is always spontaneous.

Who is "they"? As far as I know, the only one of these that is always spontaneous is a negative free energy change. A negative enthalpy change can be offset by a larger negative entropy change, and a positive entropy change can be offset by a larger positive enthalpy change.

[juanrga]

I came across a true and false question saying "A positive entropy change will always be spontaneous". I believe this is true but that leads to another question. They say that negative enthalpy change, positive entropy change, and negative free energy change is always spontaneous. But, what if we had a really low temperature and utilizing the equation delta G= delta H- TdeltaS, we could have a positive delta H, positive delta S, and then a positive delta G. Therefore, the delta G is saying that the reaction is not spontaneous while the entropy change is saying it is. What value should I ultimately look at to determine whether or not this is a spontaneous reaction? Thank you.

Entropy is the more general thermodynamic potential. A process is spontaneous when the production of entropy is positive. For an isolated system, this means :delta: S > 0.

Enthalpy and Gibbs energy are not so general. Both are thermodynamic potentials under restricted experimental constraints. For example G is a thermodynamic potential for T and p constant. Most chemical reactions of interest happen at T and p constant, that is the reason of why G is so important for chemists.

Moreover, the spontaneity criteria for H and G is a consequence of applying the constraints and the spontaneity criteria for S. That is, if S predicts that some process is spontaneous, then H and G must agree. If you are interested the demonstration for :delta: G < 0 can be given here.

[fledarmus]

Entropy is the more general thermodynamic potential. A process is spontaneous when the production of entropy is positive. For an isolated system, this means :delta: S > 0.

I have to take issue with this point. If  :delta:S is always >0 for a spontaneous process, how do you account for crystallizations or condensations? The entropy for a crystalline substance is far higher than the entropy of a liquid, and the entropy of a liquid is far higher than the entropy of a gas. For a gas to condense or for a liquid to solidify, the loss of entropy (- :delta:S) must be offset by an increase in enthalpy such that the Gibb's free energy for the process is negative. It is the Gibb's free energy  :delta:G that tells you whether a process is spontaneous.

At least, that is my understanding. I would be happy to be corrected if I am mistaken.

[juanrga]

Entropy is the more general thermodynamic potential. A process is spontaneous when the production of entropy is positive. For an isolated system, this means :delta: S > 0.

I have to take issue with this point. If  :delta:S is always >0 for a spontaneous process, how do you account for crystallizations or condensations? The entropy for a crystalline substance is far higher than the entropy of a liquid, and the entropy of a liquid is far higher than the entropy of a gas. For a gas to condense or for a liquid to solidify, the loss of entropy (- :delta:S) must be offset by an increase in enthalpy such that the Gibb's free energy for the process is negative. It is the Gibb's free energy  :delta:G that tells you whether a process is spontaneous.

At least, that is my understanding. I would be happy to be corrected if I am mistaken.

Note that in the part that you quote I say that :delta:S is always positive for isolated systems. For non-isolated systems it can be negative or zero.

The general balance law for entropy for any integral process is
$$\Delta S = \Delta_i S + \Delta_e S$$
The term ##\Delta_i S## accounts for production of entropy due to dissipative processes inside the system and the term ##\Delta_e S## accounts for the flow of entropy with the surrounds. Thermodynamics states that, with independence of boundaries ##\Delta_i S \geq 0## with the inequality holding for spontaneous processes.

If your system is isolated ##\Delta_e S = 0## and
$$\Delta S = \Delta_i S \geq 0$$

That is why I said that for an isolated system a spontaneous process verifies ##\Delta S > 0##. If the system is not isolated, this is not true and the total variation of entropy ##\Delta S ## can be positive, negative or zero depending of the flow term ##\Delta_e S##.

Precisely the special spontaneity criteria ##\Delta G < 0## for some spontaneous processes is derived from the general condition ##\Delta_i S > 0## for any spontaneous process.

I will write the proof in a subsequent message.

[juanrga]

This proof can be found in chapter 5 of the textbook «Modern Thermodynamics, from heat engines to dissipative structures» by Dilip Kondepudi and Ilya Prigogine.

We start with the definition of Gibbs energy function
$$G \equiv U + pV - TS$$
and compute its variation
$$\Delta G = \Delta U + p \Delta V + V \Delta p - T \Delta S - S \Delta T$$
Next we use the general balance law for entropy
$$\Delta S = \Delta_i S + \Delta_e S$$
and obtain
$$\Delta G = \Delta U + p \Delta V + V \Delta p - T \Delta_i S - T \Delta_e S - S \Delta T$$
For a closed system at constant temperature and pressure, ##\Delta_e S = Q/T## and ##\Delta U = Q - p \Delta V##. Substituting all back
$$\Delta G = - T \Delta_i S$$
since for any spontaneous process ##\Delta_i S > 0## it follows that
$$\Delta G < O$$
Note that entropy is always a thermodynamic potential, whereas the Gibbs energy is only a thermodynamic potential for constant pressure and temperature.

If you cannot access the above textbook, a similar proof is given in chapter 7 of the 7th Edition of «Chemical Thermodynamics: Basic Theory and Methods» by Irving M. Klotz and Robert M. Rosenberg.

Klotz and Rosenberg proof do not use the modern thermodynamics formalism used here and although they emphasize that ##\Delta G < O## provides the criteria of spontaneity for temperature and pressure constant (see their equation 7.18), they fail to say that the system has to be closed as well.