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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 $\Delta Q$ will be always smaller than (or equal to) $T\,\Delta S$ and $\Delta H=\Delta Q$, then looking at the first formula again it seems that $\Delta G$ 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. $\endgroup$ – Abhigyan Chattopadhyay 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, $\Delta G$ can only be negative 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 positive.

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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What am I doing wrong here?

You are missing the possibility of non-PV work.

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

$\Delta H$ is just the same as $\Delta Q$ under the following conditions: You are working at constant pressure, and the only work is expansion work. In that case, the change in entropy of the surrounding and the change in entropy of the system are described by $-\frac{\Delta Q}{T}$ and $\Delta S$ (assuming the latter refers to the system). Under these conditions, the reaction or process will always go in a direction that decreases $G$, i.e. $\Delta G$ will be negative.

However, it can be positive if you do work on the system, e.g. charging a battery. So it is not a general statement that $\Delta G$ can never be positive. This is different from the absolute statement about increasing entropy, as described by the second law.

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