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$$\frac{-\Delta G}{T}=\Delta S_{universe}$$

(This equation applies under isobaric and isothermal conditions.)

I understand that if $\Delta G$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

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  • $\begingroup$ Can you give an example of such are reaction? I think I might know where you confusion is but I'm not entirely sure. $\endgroup$ – bon Apr 19 '16 at 13:12
  • $\begingroup$ @bon How about an electrolysis reaction? I don’t know; I’m just thinking about nonspontaneous reactions in general... $\endgroup$ – lightweaver Apr 19 '16 at 13:14
  • $\begingroup$ In electrolysis you are putting in energy to drive the reaction forward. That energy has to come from somewhere and the process that creates in has a large, negative $\Delta G$. $\endgroup$ – bon Apr 19 '16 at 13:17
  • $\begingroup$ @bon I understand that, so that’s not really the focus of my question. I’m confused about the formula I posed above, really. My central confusion is that if you’re adding energy, then you’re increasing free energy, but you can’t decrease the universe’s entropy! $\endgroup$ – lightweaver Apr 19 '16 at 13:23
  • $\begingroup$ @lightweaver: The equation is true only when non-compression work is zero. $\endgroup$ – user5764 Apr 19 '16 at 14:48
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When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated.

The electrolysis process is not non-spontaneous. Yes, it would not occur in the absence of the battery. But the fact that it does occur with the battery there must tell you that overall it is spontaneous. The fact that you are coupling a strongly spontaneous reaction (discharge of the batter) with a non-spontaneous reaction (the electrolysis) means that overall, the electrolytic process is spontaneous.

It's incorrect and naive to treat the electrolysis as if it occurs without the battery.

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  • $\begingroup$ So you’re saying that when we’re adding energy to drive a nonspontaneous reaction, the equation for $\Delta G$ changes to incorporate the source of energy into that system? $\endgroup$ – lightweaver Apr 20 '16 at 10:46
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    $\begingroup$ You could look at it that way. To be more precise, the "system" that you are talking about should really include the battery as well. The equations follow. $\endgroup$ – orthocresol♦ Apr 20 '16 at 10:48
  • $\begingroup$ So, just to verify your answer, in order for reactions that decrease entropy to occur, we need to supply them with energy. By doing that we increase entropy much more somewhere else and thus the overall entropy of the system is increased? $\endgroup$ – Αντώνιος Κελεσίδης Aug 16 '16 at 15:49
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    $\begingroup$ @ΚελεσίδηςΑντώνιος That depends on your definition of "system". The Second Law only applies to an isolated system. If you consider the reaction vessel to be the "system" (in this case a closed system, not an isolated system), and everything outside the reaction vessel to be the "surroundings", then the system and its surroundings make up an isolated system, and the overall entropy of the isolated system increases. However the entropy of the reaction vessel, a closed system, decreases. The entropy of the surroundings increases. $\endgroup$ – orthocresol♦ Aug 16 '16 at 17:46
  • $\begingroup$ So the system, in which the entropy must and is increased, consists of the beaker with the reactants and the enviroment that supplies the necessary energy. I understand now. Thank you. $\endgroup$ – Αντώνιος Κελεσίδης Aug 16 '16 at 18:40
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Any reaction depends on two factors: enthalpy and entropy, and the summation of which is the change of the Gibbs energy by equation (for isobaric and isothermal process):

$$\Delta G = \Delta H - T \Delta S \qquad(1)$$ or, after manipulations we have

$$\frac{\Delta H-\Delta G}{T} = \Delta S \qquad(2)$$

The difference between equations (2) and one you've represented is the last one describes process of the hypothetical system (or real -> Universe) which always become to the initial state and there's no change of enthalpy: $$\Delta H = 0 \qquad(3)$$

If you will try to change Gibbs energy: $$\Delta G = G_2 - G_1 \qquad (4)$$ where$$G_i = H_i - TS_i \qquad (5)$$ or $$ G_i = U_i + PV_i - TS_i\qquad (6)$$ you have to change internal energy or volume of the system which will effect on enthalpy and the change of it: $$\Delta H \neq 0 \qquad(7)$$ and you will get the equation (2)

Answering on your question: handle the change of enthalpy (7) of the reaction.

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  • $\begingroup$ -1 No... the equation given in the question involves the entropy change of the universe aka entropy change of system + surroundings. Your equation only involves the entropy change of the system. $\endgroup$ – orthocresol♦ Apr 19 '16 at 17:39
  • $\begingroup$ That's right, this equation for the system, but how do you think where gathered heat/energy (increasing G1 increases H1 and in our case change of enthalpy will be positive) will go? Dissipation of energy. $\endgroup$ – Nikola Rostov Apr 19 '16 at 18:21
  • $\begingroup$ You should edit that into the answer, I think it is the missing link. $\endgroup$ – orthocresol♦ Apr 19 '16 at 19:04

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