Suppose you've got some reaction that has a positive $\Delta H$, and also has positive $\Delta S_\mathrm{sys}$. Due to the equation $$\Delta S_\mathrm{tot} = \Delta S_\mathrm{sys} - \frac{\Delta H}{T}$$ a low enough temperature will cause $\Delta S_\mathrm{tot}$ to be negative. At this point the reaction won't happen anymore because second law and whatnot.

What exactly is happening at the molecular level that prevents molecules from reacting with each other? The orientation of the molecules is more or less random, and the Maxwell–Boltzmann distribution means there ought to be some particles with sufficient kinetic energy, so what's happening? If the reactants are left long enough, 2 molecules with acceptable orientations & kinetic energies should whack into each other.

A while ago I found the "micro-states per macro-state" explanation of entropy. I get that entropy isn't a property of a single particle, or even of a specific arrangement (microstate) of particles (please correct if wrong). Honestly this just makes me even more clueless as to my above question.

I hope I haven't made this too unclear.

  • $\begingroup$ If it takes a trillion years for such a process to happen, do you still think of it as "happening"? $\endgroup$ – Zhe Mar 15 '19 at 22:45
  • $\begingroup$ Pretty much. It's random, so it could happen within 10 seconds, it just probably won't. Shouldn't 2nd law mean it would never even be possible for it to happen, under any circumstance? $\endgroup$ – Hal Gee Mar 15 '19 at 22:58
  • $\begingroup$ Your equation assumes that this is occuring at constant P. Your formula for $\Delta S_{tot}$ is not a general one, but a specific one. Entropy is usually defined in terms of heat. Heat is only equivalent to enthalpy when enthalpy is a valid. In textbooks we always end up using enthalpy when we have constant T and P, and we use internal energy U, when we are given constant T and V. This is because we are also given at the back of the textbook tables so we can look up U and H for given T and V or P respectively. You will have better luck understanding thermo if you use fully general equations. $\endgroup$ – Charlie Crown Mar 16 '19 at 1:31
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    $\begingroup$ If you think that, then that would be a misunderstanding. Positive entropy change doesn't mean impossible. It just mean very unlikely. $\endgroup$ – Zhe Mar 16 '19 at 1:38

To start, it's important to differentiate kinetics (which describe the rate at which reactions happen) and thermodynamics (which describe what net reaction can happen, even if it doesn't happen on the time scale of the universe's existence).

Your equations are all thermodynamic equations, so let's start there. I'm going to assume that you meant standard enthalpy and entropy. Then if we multiply through by $-T$, we end up with the definition of standard Gibbs free energy:

$\Delta G^\circ = -T\Delta S_{tot}=\Delta H^\circ_{sys} - T\Delta S^\circ_{sys}$.

If $\Delta G^\circ<0$, we say the reaction is spontaneous, but only in the thermodynamic sense that it can happen given infinite time. An important point is that this statement refers to the net reaction involved. The reverse reaction (which necessarily has a positive $\Delta G^\circ$) can happen, but always at a rate less than that of the forward reaction, such that the net reaction observed is the forward reaction. Because we're talking about thermodynamics and not kinetics, we can't make any conclusion about what those rates will be in absolute terms, only that the forward will be relatively faster than the reverse.

A different way of saying that is that if we set up a situation in which all reactants and products are at standard states (generally defined as 1 M solutes, 1 bar gases, pure solids, pure liquids), the concentration changes over time will be increases in products and decreases in reactants until the reaction reaches equilibrium. For a reaction that reaches equilibrium quickly, both the forward and reverse reaction happen quickly, but the forward slightly more quickly. At equilibrium, both reactions are still occurring, but now at exactly the same rate.

For a concrete example, consider the dissolution of a soluble compound like NaCl. If we have a solution of 1 M NaCl in water and we add excess solid NaCl, more NaCl will dissolve, but even when the solution becomes saturated (equilibrium), ions are constantly leaving the solid and going into solution while other ions are depositing from solution to the solid at the same rate.

In your hypothetical case where a change in T results in $\Delta G^\circ$ becoming positive, the reactions again are still happening. The difference is just that the reverse is faster than the forward when things are at standard states. It turns out that dissolution of NaCl is one of those cases you described, where $\Delta H^\circ >0$ and $\Delta S^\circ>0$. That means that if you made a 1 M solution and then cooled it, at a certain point solid would precipitate out, meaning that you've gotten to the point where at the standard state, the reverse reaction is faster than the forward reaction. At this temp, $\Delta G^\circ$ for dissolution is positive. If you stay at that temp, you can wait infinite time, and the solid will never fully redissolve.

All of that said, it sounds like your question might really be about kinetics, that is, how much can you slow down a reaction by cooling it? That question has little to do with entropy and instead relates to activation energies.

  • $\begingroup$ So for the cold NaCl solution, entropy change of dissolution would be negative. But in a really, really small time frame, an ion might come off the lattice. In the next moment, sure, 2 ions may rejoin, but in the first moment the overall entropy change is negative. Doesn't this violate 2nd law? $\endgroup$ – Hal Gee Mar 16 '19 at 10:18
  • $\begingroup$ @Hal Gee you are just looking at those two ions neglecting what happens to those already in solution. If they were alone, their going in solution doesn't violate anything. $\endgroup$ – Alchimista Mar 16 '19 at 10:58
  • $\begingroup$ Ok, but if you pick a moment during which the only thing that is happening to the lattice is an ion coming off, isn't this a negative entropy change? The rest of the ions in solution during that moment aren't doing anything, so they shouldn't make a difference. $\endgroup$ – Hal Gee Mar 16 '19 at 13:22
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    $\begingroup$ The second law is a statement of the statistical behavior of a large ensemble over time. It does not apply to a momentary fluctuation of a single atom. $\endgroup$ – Andrew Mar 16 '19 at 16:24

Assume the entropy change of the system associated with a reaction is independent of temperature and positive, as you stated. Assume also that the reaction is endothermic, with constant enthalpy change, as you stated. Finally say that you lower T so much that
$$S_\mathrm{tot} = \Delta S_\mathrm{sys} - \frac{\Delta H}{T}=\Delta S_\mathrm{sys}+\Delta S_\mathrm{surr}<0$$ The reason a reaction stops occurring (is no longer spontaneous) is that you need to transfer heat from the surroundings to the system in order to have enough energy to make the required change to the system (this is the meaning of "endothermic"). However, the loss of entropy of the surroundings associated with this heat transfer would exceed the gain in entropy of the system during the course of the transformation. Now, you can simply say that "this violates the second law of thermodynamics": you cannot spontaneously take heat away from (reduce the entropy of) the surroundings without a correspondingly equal or greater increase in the entropy of the system. If this sounds tautological, it is because it is a fundamental law. The proof is in the fact that the opposite has never been observed.


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