0
$\begingroup$

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?

|improve this question
$\endgroup$
  • $\begingroup$ What is $\Delta Q$ in this case? $\endgroup$ – Erik Kjellgren May 28 '18 at 17:42
  • 2
    $\begingroup$ 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
0
$\begingroup$

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.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.