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I have a question regarding this formula for Gibbs free energy of a system: $\Delta$G = $\Delta$H - T$\Delta$S.

According to the second law of thermodynamics: T$\Delta$S $\ge$ $\Delta$Q and to my understanding, $\Delta$H is just the same as $\Delta$Q.

Now, if ΔQ will be always smaller than (or equal to) TΔS and ΔH = ΔQ, then looking at the first formula again it seems that ΔG can only be smaller than (or equal to) zero.

What am I doing wrong here?

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  • $\begingroup$ What is $\Delta Q$ in this case? $\endgroup$ – Erik Kjellgren May 28 '18 at 17:42
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    $\begingroup$ Who says $\Delta H$ is always equal to Q? $\endgroup$ – Chet Miller May 28 '18 at 18:01
  • $\begingroup$ @ChesterMiller ∆H is undeniably not always equal to Q, but at an elementary level, which the OP seems to be, that approximation is used in a valid manner to explain it with ease to students, who can later learn more. It's like laying a temporary groundwork to be filled later, but at least as long as they're dealing with isobaric consitions, ∆H=Q. $\endgroup$ – AbhigyanC May 28 '18 at 18:12
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You didn't do anything wrong here... This was a very good observation. This is one of the criteria of spontaneity, which I believe you haven't learnt yet.

As you correctly wrote, ∆G can only be -ve or 0 for any reaction. This is actually a consequence of the second law of thermodynamics, which also goes on to say that in any spontaneous reaction the entropy change of the universe is always +ve.

Essentially, the answer to your question heading is: in any spontaneous reaction, the Gibb's free energy change can never be positive. Since you took the conditions of the second law, which essentially explains spontaneity, you arrived at this result.

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